LIBRARY 

OF  THE 

University  of  California. 


Class 


SCHOOL    ALGEBRA. 


BY 

G.    A.    WENTWORTH,   A.M., 

Author  or  a  Seribs  of  Text-Books  in  Mathematics. 


V     or  THF  \ 

UNIVERSITY    I 


of 

irc 


>XKc 


BOSTON,  U.S.A.: 

PUBLISHED   BY   GIISTN   &   COMPANY. 

1896. 


Entered,  according  to  Act  of  Congress,  in  the  year  1890,  by 

G.  A.   WENT  WORTH, 

in  the  Office  of  the  Librarian  of  Congress,  at  Washington. 


All  Rights  Reserved. 


Ttpography  bt  J.  S.  Gushing  &  Co.,  Boston,  U.S.A. 


Prbsswork  by  Ginn  85  Co.,  Boston,  U.S.A. 


PREFACE. 


This  book,  as  the  name  implies,  is  written  for  High  Schools  and 
Academies,  and  is  a  thorough  and  practical  treatment  of  the  prin- 
ciples of  Elementary  Algebra,  It  covers  sufficient  ground  for 
admission  to  any  American  college,  and  with  the  author's  Col- 
lege Algebra  makes  as  extended  a  course  as  the  time  allotted 
to  this  study  in  our  best  schools  and  colleges  will  allow.  Great 
care  has  been  taken  to  present  the  best  methods,  so  that  students 
in  going  from  the  lower  book  to  the  higher  will  have  a  good 
foundation,  and  have  .nothing  to  unlearn. 

The  problems  are  carefully  graded.  They  are  for  the  most  part 
new ;  either  original  or  selected  from  recent  examination  papers. 
They  are  sufficiently  varied  and  interesting,  and  are  not  so  difficult 
as  to  discourage  the  beginner.  The  early  chapters  are  quite  full; 
for  even  if  a  student  is  perfectly  familiar  with  the  operations  of 
Arithmetic,  he  must  have  time  to  learn  the  language  and  the 
fundamental  processes  of  Algebra. 

The  introductory  chapter  should  be  read  and  discussed  in  the 
recitation  room.  This  chapter  brings  before  the  student  in  brief 
revieW  the  knowledge  he  has  already  gained  from  the  study  of 
Arithmetic,  states  and  proves  the  general  laws  of  numbers,  sets 
forth  clearly  the  advantage  of  using  letters  to  represent  numbers 
in  the  statement  of  general  laws,  and  leads  him  to  see  at  the 
outset  that  Algebra,  like  Arithmetic,  treats  of  numbers.  In  this 
chapter,  also,  the  meaning  of  negative  quantities  is  explained,  and 
the  laws  which  regulate  the  combinations  of  diflferent  arithmetical 
numbers  are  shown  to  apply  to   algebraic   numbers.     It  is  hoped 


IV  PREFACE. 

that  a  free  discussion  of  these  elementary  principles  will  do  much 
to  prevent  that  vagueness  which  the  beginner  invariably  experi- 
ences if  he  fails  to  connect  the  laws  of  Algebra  with  what  he  has 
learned  in  Arithmetic. 

Answers  to  the  problems  are  bound  separately,  in  paper  covers, 
and  will  be  furnished  free  to  pupils  when  teachers  apply  to  the 
publishers  for  them. 

Any  corrections  or  suggestions  relating  to  the  work  will  be 
thankfully  received. 

G.  A.  WENTWORTH. 

EXETEE,    N.H., 

June,  1890. 


CONTENTS. 


Chaptkb  Paos 

I.     Introduction 1 

II.    Addition  and  Subtraction 28 

III.  Multiplication 39 

IV.  Division 48 

V.    Simple  Equations    ....              59 

VI.     Multiplication  and  Division 74 

VII.     Factors 84 

VIII.     Common  Factors  and  Multiples 105 

IX.     Fractions    .     ., 120 

X.     Fractional  Equations 144 

XI.  Simultaneous  Equations  of  the  First  Degree    .  163 

XII.  Problems  involving  Two  Unknown  Numbers  .    .179 

XIII.  Inequalities 197 

XIV.  Involution  and  Evolution 200 

XV.     Theory  of  Exponents 217 

XVI.     Radical  Expressions 225 

XVII.     Imaginary  Expressions 243 

XVIII.  -^  Quadratic  Equations 249 

XIX.    Simultaneous  Quadratic  Equations 277 

XX.     Properties  of  Quadratics 287 

—  XXI.     Ratio,  Proportion,  and  Variation 293 

XXII.     Progressions 310 

XXIII.  Properties  of  Series  . 323 

XXIV.  Binomial  Theorem 330 

XXV.     Logarithms 342 

XXVI.     General  Review  Exercise.         356 


) 


'^     or  THF  ^ 

UNIVERSITY  ) 

OF  J 

SCHOOL  ALGEBRA. 


CHAPTER  I. 
INTRODUCTION. 

1.  Units.  In  counting  separate  objects  the  standards  by 
which  we  count  are  called  units ;  and  in  measuring  contin- 
uous magnitudes  the  standards  by  which  we  measure  are 
called  units.  Thus,  in  counting  the  boys  in  a  school,  the 
unit  is  a  boy ;  in  selling  eggs  by  the  dozen,  the  unit  is  a 
dozen  eggs  ;  in  selling  cloth  by  the  yard,  the  unit  is  a  yard 
of  cloth  ;  in  measuring  short  distances,  the  unit  is  an  inch, 
a  foot,  or  a  yard  ;  in  measuring  long  distances,  the  unit  is 
a  rod  or  a  mile. 

2.  Numbers.  Repetitions  of  the  unit  are  expressed  by 
numbers.  If  a  man,  in  sawing  logs  into  boards,  wishes  to 
keep  a  count  of  the  logs,  he  makes  a  straight  mark  for 
every  log  sawed,  and  his  record  at  different  times  will  be 
as  follows : 

/    //    ///    ////    iHj     m/ / 
tHj  II    nu  III    tw  nil    fuj  rw 

These  representative  groups  are  named  one,  two,  three, 
four,  five,  six,  seven,  eight,  nine,  ten,  etc.,  and  are  known 
collectively  under  the  general  name  of  numbers.  It  is 
obvious  that  these  representative  groups  will  have  the 
same  meaning,  whatever  the  nature  of  the  unit  counted. 


2  SCHOOL   ALGEBRA. 

3.  Quantities.  The  word  "  quantity "  (from  the  Latin 
quantus,  how  much)  implies  both  a  unit  and  a  number. 
Thus,  if  we  inquire  how  much  wheat  a  bin  will  hold,  we 
mean  how  many  bushels  of  wheat  it  will  hold.  If  we 
inquire  how  much  carpeting  there  is  in  a  certain  roll, 
we  mean  how  many  yards  of  carpeting.  If  we  inquire 
how  much  wood  there  is  on  a  certain  wood-lot,  we  mean 
how  many  cords  of  wood. 

4.  Number-Symbols  in  Arithmetic.  Instead  of  groups  of 
straight  marks,  we  use  in  Arithmetic  the  arbitrary  sym- 
bols 1,  2,  3,  4,  5,  6,  7,  8,  9,  called  figures,  for  the  numbers 
one,  two,  three,  four,  five,  six,  seven,  eight,  nine. 

The  next  number,  ten,  is  indicated  by  writing  the  figure 
1  in  a  different  position,  so  that  it  shall  signify  not  one,  but 
ten.  This  change  of  position  is  effected  by  introducing  a 
new  symbol,  0,  called  nought  or  zero,  and  signifying  none. 
Thus,  in  the  symbol  10,  the  figure  1  occupying  the  second 
place  from  the  right,  signifies  a  collection  of  ten  things,  and 
the  zero  signifies  that  there  are  no  single  things  over.  The 
symbol  11  denotes  a  collection  of  ten  things  and  one  thing 
besides.  All  succeeding  numbers  up  to  the  number  con- 
sisting of  10  tens  are  expressed  by  writing  the  figure  for 
the  number  of  tens  they  contain  in  the  second  place  from 
the  right,  and  the  figtire  for  the  number  of  units  besides  in 
the  first  place.  The  number  consisting  of  10  tens  is  called 
a  hundred,  and  the  hundreds  of  a  number  are  written  in  the 
third  place  from  the  right.  The  number  consisting  of  10 
hundreds  is  called  a  thousand,  and  the  thousands  are  writ- 
ten in  the  fourth  place  from  the  right;  and  so  on, 

5.  The  Natural  Series  of  Numbers.  Beginning  with  the 
number  one,  each  succeeding  number  is  obtained  by  put- 
ting one  more  with  the  preceding  number.    If  from  a  given 


INTRODUCTION.  8 

point  marked  0,  we  draw  a  straight  line  to  the  right,  and 
beginning  from  this  point  lay  off  units  of  length,  the  suc- 
cessive repetitions  of  the  unit  will  be  denoted  by  the  natural 
series  of  numbers  1,  2,  3,  4,  etc.     Thus, 

12       3       4       5        6       7       etc. 


6.  The  reader  will  notice  that  number  symbols  in  Arith- 
metic stand  for  particular  numbers,  and  that  these  symbols 
indicate  a  method  of  making  up  the  number,  but  not  neces- 
sarily the  method  by  which  the  number  is  actually  made 
up.  Thus,  if  a  man  has  66  dollars  in  bank-notes,  he  may 
have,  as  the  number  66  indicates,  6  ten-dollar  bills  and  6 
one-dollar  bills,  but  this  is  not  the  only  way  in  which  the 
66  dollars  may  be  made  up. 

7.  Integral  and  Fractional  Numbers.  When  the  things 
counted  are  whole  units,  the  numbers  which  count  them 
are  called  whole  numbers,  integral  numbers,  or  integers,  where 
the  adjective  is  transferred  from  the  things  counted  to  the 
numbers  which  count  them.  But  if  the  things  counted  are 
only  parts  of  units,  the  numbers  which  count  them  are 
called  fractional  numbers,  or  simply  fractions,  where  again 
the  adjective  is  transferred  from  the  things  counted  to  the 
numbers  which  count  them. 

To  represent  the  parts  of  a  given  unit,  two  number- 
symbols  are  used,  one  to  name  the  parts  into  which  the 
unit  is  divided,  and  therefore  called  the  denominator,  and 
the  other  to  denote  the  number  of  parts  taken,  and  there- 
fore called  the  numerator.  The  denominator  is  written 
below  the  numerator  with  a  line  between  them.  Thus,  in 
the  fraction  |-  the  9  shows  that  the  unit  is  divided  into 
nine  equal  parts,  called  ninths  of  the  unit,  and  the  7  shows 
that  seven  of  these  equal  parts  are  taken. 


/ 

4  SCHOOL   ALGEBRA. 

8.  Principal  Signs  of  Operations.     The  sign  +,  read  'plus, 

indicates  that  the  number  after  the  sign  is  to  be  added  to 
the  number  before  the  sign.  Thus,  5  +  4  means  that  4 
is  to  be  added  to  5. 

The  sign  — ,  read  minus,  indicates  that  the  number  after 
the  sign  is  to  be  subtracted  from  the  number  before  the  sign. 
Thus,  8  —  4  means  that  4  is  to  be  subtracted  from  8. 

The  sign  X,  read  ^mes,  indicates  that  the  number  after 
the  sign  is  to  be  multiplied  by  the  number  before  the  sign. 
Thus,  5x4  means  that  4  is  to  be  multiplied  by  5. 

The  sign  -^-,  read  divided  by,  indicates  that  the  number 
before  the  sign  is  to  be  divided  by  the  number  after  the 
sign.     Thus,  8-^-4  means  that  8  is  to  be  divided  by  4. 

The  operation  of  division  is  also  indicated  by  placing 
the  dividend  over  the  divisor  with  a  line  between  them. 
Thus,  -|  means  the  same  as  8-r-4. 

9.  Signs  of  Eelation.  The  sign  =,  read  equals,  or  is  equal 
to,  when  placed  between  two  numbers,  indicates  that  they  are 
equal.    Thus,  8  +  4  =  12  means  that  8+4  is  the  same  as  12. 

The  sign  >,  read  is  greater  than,  indicates  that  the  num- 
ber which  precedes  the  sign  is  greater  than  the  number 
which  follows  it.  Thus,  8  +  4  >  10  means  that  8  +  4  is 
greater  than  10. 

The  sign  <,  read  is  less  than,  indicates  that  the  number 
which  precedes  the  sign  is  less  than  the  number  which  fol- 
lows it.     Thus,  8  +  4<16  means  that  8+4  is  less  than  16. 

10.  Signs  of  Deduction  and  of  Continuation.  The  sign  .'. 
stands  for  the  word  "therefore"  or  "hence."  The  sign 
or stands  for  the  words  "  and  so  on." 

11.  Number-Symbols  in  Algebra.  Algebra,  like  Arith- 
metic, treats  of  numbers,  and  employs  the  letters  of  the 
alphabet  in  addition  to  the  figures  of  Arithmetic  to  represent 


INTRODUCTION.  0 

numbers.  The  letters  of  the  alphabet  are  used  as  general 
symbols  of  numbers  to  which  any  particular  values  may  be 
assigned.  In  any  particular  problem,  however,  a  letter  must 
be  supposed  to  have  the  same  particular  value  throughout 
the  investigation  or  discussion  of  the  problem. 

These  general  symbols  are  of  great  advantage  in  investi- 
gating and  stating  general  laws ;  in  exhibiting  the  actual 
method  in  which  a  number  is  made  up ;  and  in  represent- 
ing unknown  numbers  which  are  to  be  discovered  from  their' 
relations  to  known  numbers. 

The  advantage  of  representing  numbers  by  letters  .will 
be  more  clearly  seen  later  on.  For  the  present  it  will  be 
sufficient  for  the  beginner  to  understand  that  every  letter, 
and  every  combination  of  letters,  and  every  combination  of 
figures  and  letters  used  in  Algebra,  represents  some  number. 
Thus,  the  number  of  dollars  in  a  package  of  bank-notes 
can  be  represented  by  x ;  but  if  the  package  consists  of  ten- 
dollar  bills,  five-dollar  bills,  two-dollar  bills,  and  one-dollar 
bills,  and  if  we  denote  the  number  of  ten-dollar  bills  by  a, 
of  five-dollar  bills  by  b,  of  two-dollar  bills  by  c,  and  of  one- 
dollar  bills  by  d,  the  whole  number  of  dollars  in  the  package 
will  be  represented  by  10a-{- bb  -{-2c-{- d. 

In  this  particular  case  x  and  lOa-j-  bb  -\-2c-{-  d  both 
stand  for  the  same  number. 

12.  Substitution.  It  is  obvious  that  the  same  operation 
on  each  of  the  above  expressions  will  produce  results  that 
agree  in  value,  and  therefore  that  either  may  be  substituted 
for  the  other  at  pleasure.     In  short, 

Every  algebraic  expression  represents  some  number,  and 
may  be  operated  upon  as  if  it  were  a  single  symbol  standing 
for  the  number  which  it  represents. 

13.  Factors.  When  a  number  consists  of  the  product  of  two 
or  more  numbers,  each  of  these  numbers  is  called  a  factor  of 


6  SCHOOL    ALGEBRA. 

the  product.     If  these  numbers  are  denoted  by  letters,  the 
sign  X  is  omitted.     Thus,  instead  oi  axb,  we  write  ab. 

14.  Coefficients.  A  known  factor  prefixed  to  another 
factor  to  show  how  many  times  that  factor  is  taken  is  called 
a  coefficient.      Thus,  in7x  the  factor  7  is  the  coefficient  of  x. 

15.  Powers.  A  product  consisting  of  two  or  more  equal 
factors  is  called  a  power  of  that  factor. 

The  index  or  exponent  of  a  power  is  a  number-symbol 
placed  at  the  right  of  a  number,  to  show  how  many  times 
the  number  is  taken  as  a  factor.  Thus,  2*  is  written  instead 
of  2x2x2x2;  a^  instead  of  aaa. 

The  second  power  of  a  number  is  generally  called  the 
square  of  that  number ;  the  third  power  of  a  number,  the 
cube  of  that  number. 

16.  Eoots.  The  root  of  a  number  is  one  of  the  equal  fac- 
tors of  that  number ;  the  square  root  of  a  number  is  one 
of  the  iivo  equal  factors  of  that  number ;  the  cube  root  of  a 
number  is  one  of  the  three  equal  factors  of  that  number ; 
and  so  on.  The  sign  V,  called  the  radical  sign,  indicates 
that  a  root  is  to  be  found.  Thus,  Vl,  or  V4,  means  that 
the  square  root  of  4  is  to  be  taken  ;  VS  means  that  the 
cube  root  of  8  is  to  be  taken ;  and  so  on. 

The  number-symbol  written  above  the  radical  sign  is 
called  the  index  of  the  root. 

17.  An  algebraic  expression  is  a  number  written  with  alge- 
braic symbols  ;  an  algebraic  expression  consists  of  one  sym- 
bol, or  of  several  symbols  connected  by  signs  of  operation. 

A  term  is  an  algebraic  expression  the  parts  of  which  are 
not  separated  by  the  sign  of  addition  or  subtraction.  Thus, 
Sab,  bxX  4y,  Sab -^4^X7/  are  terms. 


INTRODUCTION.  7 

A  simple  expression  is  an  expression  of  one  term. 
A  compound  expression  is  an  expression  of  two  or  more 
terms. 

18.  Positive  and  Negative  Terms.  The  terms  of  a  com- 
pound expression  preceded  by  the  sign  +  are  called  posi- 
tive terms,  and  the  terms  preceded  by  the  sign  —  are  called 
negative  terms.    The  sign  -f  before  the  first  term  is  omitted. 

19.  Parentheses.  If  a  compound  expression  is  to  be 
treated  as  a  whole  it  is  enclosed  in  a  parenthesis.  Thus, 
2  X  (10  +  5)  means  that  we  are  to  add  5  to  10  and  multiply 
the  result  by  2 ;  if  we  were  to  omit  the  parenthesis  and 
write  2  X  10  +  5,  the  meaning  would  be  that  we  were  to 
multiply  10  by  2  and  add  5  to  the  result. 

Instead  of  parentheses,  we  use  with  the  same  meaning 
brackets  [  ],  braces  \  \ ,  and  a  straight  line  called  a  vinculum. 

Thus,  (5  4-  2),  [5  +  2],  ^5  +  2J,  5T2,   ,  £ '  ^^^  ^®^^  *^^* 

the  expression  5  +  2  is  to  be  treated  as  the  single  symbol  7. 

20.  Bules  for  removing  Parentheses.  If  a  man  has  10  dol- 
lars and  afterwards  collects  3  dollars  and  then  2  dollars, 
it  makes  no  difference  whether  he  adds  the  3  dollars  to  his 
10  dollars,  and  then  the  2  dollars,  or  puts  the  3  and  2 
dollars  together  and  adds  their  sum  to  his  10  dollars. 

The  first  process  is  represented  by  10  +  3  +  2. 
The  second  process  is  represented  by  10  +  (3  +  2). 

Hence  10 +  (3 +  2)  =  10  +  3  +  2.  (1) 

If  a  man  has  10  dollars  and  afterwards  collects  3  dol- 
lars and  then  pays  a  bill  of  2  dollars,  it  makes  no  differ- 
ence whether  he  adds  the  3  .dollars  collected  to  his  10 
dollars  and  pays  out  of  this  sum  his  bill  of  2  dollars,  or 
pays  the  2  dollars  from  the  3  dollars  collected  and  adds 
the  remainder  to  his  10  dollars. 


8  SCHOOL    ALGEBRA. 

The  first  process  is  represented  by  10  +  3  —  2. 
The  second  process  is  represented  by  10  +  (3  —  2). 

Hence  10  +  (3-2)=  10  +  3-2.  \  (2) 

From  (1)  and  (2)  it  follows  that  if  a  compound  expres- 
sion is  to  be  addedy  the  parenthesis  may  be  removed  and 
each  term  in  the  parenthesis  retain  its  prefixed  sign. 

If  a  man  has  10  dollars  and  has  to  pay  two  bills,  one  of 
3  dollars  and  one  of  2  dollars,  it  makes  no  difference  whether 
he  takes  3  dollars  and  2  dollars  in  succession,  or  takes  the  3 
and  2  dollars  at  one  time,  from  his  10  dollars. 

The  first  process  is  represented  by  10  —  3  —  2. 

The  second  process  is  represented  by  10  —  (3  +  2). 

Hence  10-(3  +  2)  =  10-3-2.  (3) 

If  a  man  has  10  dollars  consisting  of  2  five-dollar  bills, 
and  has  a  debt  of  3  dollars  to  pay,  he  can  pay  his  debt  by 
giving  a  five-dollar  bill  and  receiving  2  dollars. 

This  process  is  represented  by  10  —  5  +  2. 

Since  the  debt  paid  is  three  dollars,  that  is,  (5  —  2)  d©l- 
lars,  the  number  of  dollars  he  has  left  can  evidently  be 
expressed  by         10  —  (5  —  2) 

Hence  10  --  (5  ^  2)  ="  10-5  +  2.  (4) 

From  (3)  and  (4)  it  follows  that  if  a  compound  expres- 
sion is  to  be  subtracted,  the  parenthesis  may  be  removed, 
provided  the  sign  before  each  term  within  the  parenthesis 
is  changed,  the  sign  +  to  — ,  and  the  sign  —  to  +. 

Exercise  1. 
Perform  the  operations  indicated,  and  simplify  . 

1.  7 +'(3 -2).      4.   5x(2  +  3).       7.    (7-3)  x  (5-2). 

2.  7-(3-2).       5.   (5  +  3)-^2.       8.   (8-2)--(5-2). 

3.  7_(3  +  2).       6.    5x(3-2).        9.    3  X  (12-6-2). 


INTRODUCTION.  9 

21.  rnndamental  Laws  of  Numbers.  We  are  so  occupied 
in  Arithmetic  with  the  application  of  numbers  to  the  ordi- 
nary problems  of  every-day  life  that  we  pay  little  attention 
to  the  investigation  of  the  fundamental  laws  of  numbers. 
It  is,  however,  very  important  that  the  beginner  in  Algebra 
should  have  clear  ideas  of  these  laws,  and  of  the  extended 
meaning  which  it  is  necessary  to  give  in  Algebra  to  cer- 
tain words  and  signs  used  in  Arithmetic  ;  and  that  he 
should  see  that  every  such  extension  of  meaning  is  con- 
sistent with  the  meaning  previously  attached  to  the  word 
or  sign,  and  with  the  general  laws  of  numbers.  We  shall, 
therefore,  give  general  definitions  for  the  fundamental  oper- 
ations upon  numbers  and  then  state  the  laws  which  apply 
to  them. 

22.  Addition.  The  process  of  finding  the  result  when 
two  or  more  numbers  are  taken  together  is  called  addition, 
and  the  result  is  called  the  sum, 

23.  Subtraction.  The  process  of  finding  the  result  when 
one  number  is  taken  from  another  is  called  subtraction,  and 
the  result  is  called  the  difference  or  remainder.  The  number 
taken  away  is  called  the  subtrahend,  and  the  number  from 
which  the  subtrahend  is  taken  is  called  the  minuend. 

In  practice  the  difference  is  found  by  discovering  the 
number  which  must  be  added  to  the  subtrahend  to  give  the 
minuend.  If  the  subtrahend  consists  of  two  or  more  terms, 
we  add  these  terms  and  then  determine  the  number  which 
must  be  added  to  their  sum  to  make  it  equal  to  the  minu- 
end. Thus,  if  a  clerk  in  a  store  sells  articles  for  10  cents, 
15  cents,  and  30  cents,  and  receives  a  dollar  bill  in  pay- 
ment, he  makes  change  by  adding  these  items  and  then 
adding  to  their  sum  enough  change  to  make  a  dollar. 

From  the  nature  of  this  process  it  is  obvious  that  the 
general  laws  of  numbers  which  apply  to  addition  apply 


10  SCHOOL   ALGEBRA. 

also  to  subtraction,  and  that  we  may  take  for  the  general 
definition  of  subtraction 

The  operation  of  finding  from  two  given  numbers,  called 
minuend  and  subtrahend,  a  third  number,  called  difference, 
which  added  to  the  subtrahend  will  give  the  minuend. 

24.  Multiplication.  The  process  of  finding  the  result 
when  a  given  number  is  taken  as  many  times  as  there  are 
units  in  another  number  is  called  multiplication,  and  the  re- 
sult is  called  the  product. 

This  definition  fails  when  the  multiplier  is  a  fraction,  for 
we  cannot  take  the  multiplicand  b.  fraction  of  a  time.  We 
therefore  consider  what  extension  of  the  meaning  of  multi- 
plication can  be  made  so  as  to  cover  the  case  in  question. 
When  we  multiply  by  a  fraction  we  divide  the  multiplicand 
into  as  many  equal  parts  as  there  are  units  in  the  denomi- 
nator and  take  as  many  of  these  parts  as  there  are  units 
in  the  numerator.  If,  for  instance,  we  multiply  8  by  f ,  we 
divide  8  into  four  equal  parts  and  take  three  of  these  parts, 
getting  6  for  the  product.  We  see  that  f  is  f  of  1,  and  6 
is  I  of  8 ;  that  is,  the  product  6  is  obtained  from  the  mul- 
tiplicand 8  precisely  as  the  multiplier  J  is  obtained  from  1. 
The  same  is  true  when  the  multiplier  is  an  integral  number. 

Thus,  in  6  X  8  =  48, 
the  multiplier  '        6isl  +  l-fl  +  l-fl  +  l, 
and  the  product    48  is  8  +  8  +  8  +  8  +  8-}- 8. 

We  may,  therefore,'  take  for  the  general  definition  of 
multiplication 

The  operation  of  finding  from  two  given  numbers,  called 
multiplicand  and  multiplier,  a  third  number  called  product, 
which  IB  formed  from  the  multiplicand  as  the  multiplier  is 
formed  from  unity. 


INTRODUCTION.  11 

26.  Division.  To  divide  48  by  8  is  to  find  the  number 
of  times  it  is  necessary  to  take  8  to  make  48.  Here  the 
product  and  one  factor  are  given  and  the  other  factor  is 
required.  We  may  therefore  take  for  the  general  definition 
of  division 

The  operation  by  which  when  the  product  and  one  factor 
are  given  the  other  factor  is  found. 

With  reference  to  this  operation  the  product  is  called 
the  dividend,  the  given  factor  the  divisor,  and  the  required 
factor  the  quotient. 

26.  The  Commutative  Law.  If  we  have  a  group  of  3 
things  and  another  group  of  4  things,  we  shall  have  a 
group  of  7  things,  whether  we  put  the  3  things  with  the 
4  things  or  the  4  things  with  the  3  things ;  that  is, 

4  +  3  =  3  +  4. 

It  is  evident  that  the  truth  of  the  above  statement  does 
not  depend  upon  the  particular  numbers  3  and  4,  but  that 
the  statement  is  true  for  any  two  numbers  whatever.  Thus, 
in  case  of  any  two  numbers  we  shall  have 

First  number  +  second  number  =  second  number  +  first 
number. 

If  we  let  a  stand  for  the  first  number  and  h  for  the  second 
number,  this  statement  may  be  written  in  the  much  shorter 

form 

a-\-  0  =  o-\-  a. 

This  is  the  commutative  law  of  addition,  and  may  be 
stated  as  follows : 


//     Additions  may  he  performed  in  any  order. 


27.   Also,  if  we  have  5  lines  of  dots  with  10  dots  in  a 
line,  the  whole  number  of  dots  will  be  expressed  by  5  X  10. 


12  SCHOOL  ALGEBRA. 


II 


If  we  consider  the  dots  as  10  columns  with  5  dots  in  a 
column,  the  number  will  be  expressed  by  10  X  5. 

That  is,  5  X  10  =  10  X  5. 

Again,  if  we  divide  a  given  length  into  6  equal  parts, 

I I 1 1 . 1 I 

one-third  of  the  line  will  contain  2  of  these  parts,  and  one- 
half  the  line  will  contain  3  of  these  parts.  Now  one-third 
of  one-half  will  be  1  of  these  parts,  and  one-half  of  one- 
third  will  be  1  of  these  parts ;  that  is. 

Therefore,  if  a  and  h  stand  for  any  two  numbers,  integral 
or  fractional,  we  shall  have 

ab  =  ha. 

This  is  the  commutative  law  of  multiplication,  and  may 
be  stated  as  follows : 

Multiplications  may  he  performed  in  any  order. 

28.  The  Distributive  Law.  The  expression  4  X  (5  -]-  3) 
means  that  we  are  to  take  the  sum  of  the  numbers  5  and  3 
four  times.  The  process  can  be  represented  by  placing  five 
dots  in  a  line,  and  a  little  to  the  right  three  more  dots  in 
the  same  line,  and  then  placing  a  second,  third,  and  fourth 
line  of  dots  underneath  the  first  line  and  exactly  similar  to 

^^*  •••••  ••• 

•  ••••  ••• 

•  ••••  ••• 

•  ••••  ••• 


INTRODUCTION.  13 

There  are  (5  +  3)  dots  in  each  line,  and  4  lines.  The 
total  number  of  dots,  therefore,  is  4  X  (5  +  3). 

We  see  that  in  the  left-hand  group  there  are  4x5  dots, 
and  in  the  right-hand  group  4x3  dots.  The  sum  of  these 
two  numbers  (4  X  5)  +  (4  X  3)  must  be  equal  to  the  total 
number ;  that  is, 

4x(5  +  3)  =  (4x5)-f-(4x3). 

Again,  the  expression  4  X  (8  —  3)  means  that  3  is  to  be 
taken  from  8,  and  the  remainder  to  be  multiplied  by  4. 
The  process  can  be  represented  by  placing  eight  dots  in  a 
line  and  crossing  the  last  three,  and  then  placing  a  second, 
third,  and  fourth  line  of  dots  underneath  the  first  line  and 
exactly  similar  to  it. 

iii 

Hi 

Hi 

Hi 

The  whole  number  of  dots  not  crossed  in  each  line  is 
evidently  (8  —  3),  and  the  whole  number  of  lines  is  4. 
Therefore  the  total  number  of  dots  not  crossed  is 

4  X  (8  -  3). 

The  total  number  of  dots  (crossed  and  not  crossed)  is 
(4  X  8),  and  the  total  number  of  dots  crossed  is  (4  X  3). 
Therefore  the  total  number  of  dots  not  crossed  is 

(4  X  8)  -  (4  X  3)  ; 
that  is,  4  X  (8  -  3)  =  (4  X  8)-(4  X  3). 

Hence,  by  the  commutative  law 

(8-3)x4  =  (8x4)-(3x4). 
In  like  manner,  if  a,  5,  c,  and  d  stand  for  any  numbers, 
we  have        .  ay.Q)-\-c- d)^al^ac- cd. 


14 


SCHOOL    ALGEBRA. 


This  is  the  distributive  law,  and  may  be  stated  as  follows: 

In  multiplying  a  coTnpound  expression  hy  a  simple  ex- 
pression the  result  is  obtained  by  multiplying  each  term  of 
the  compound  expression  by  the  simple  expression,  and  writ- 
ing down  the  successive  products  with  the  same  signs  as 
those  of  the  original  terms. 

29.  The  Associative  Law.  The  terms  of  an  expression  may 
be  grouped  in  any  manner.  For  if  we  have  several  num- 
bers to  be  added,  the  result  will  evidently  be  the  same, 
whether  we  add  the  numbers  in  succession  or  arrange  them 
in  groups  and  add  the  sums  of  these  groups.     Thus, 

a-\-b  -{-  c  -\~  d-\-  e 
=  a-\-{b-^c)-{-{d  +  e) 
=  {a  +  b)-{-{c  +  d-\-e). 

Likewise,  if  in  the  rectangular  solid  represented  in  the 
margin  we  suppose  A  JB  to  contain  5  units  of  length,  JBO  S 
units,  and  CD  7  units.  The  base  may 
be  divided  into  square  units.  There 
will  be  3  rows  of  6  square  units  each. 
Upon  each  square  unit  a  cubic  unit  may 
be  formed,  and  we  shall  have  (3x5) 
cubic  units.  Upon  these  another  tier  of 
(3  X  5)  cubic  units  may  be  formed,  and 
then  another  tier  of  the  same  number, 
and  the  process  continued  until  we  have  7  tiers  of  (3  X  5) 
cubic  units.  Hence  the  number  of  cubic  units  in  the  solid 
will  be  represented  by  7  X  (3  X  5). 

Upon  the  right-hand  square  in  the  back  row  a  pile  of  7 
cubic  units  may  be  formed,  upon  the  next  square  to  the 
left  another  pile  of  7  cubic  units  may  be  formed,  and 
upon  the  next  square  another,  and  the  process  continued 
until  we  have  a  pile  of  7  cubic  units  on  each  square  in  the 


D 


i^ 


A^^. 


//// 


^'7 

^'.7 


B 


INTRODUCTION.  15 

back  row.  We  shall  then  have  (5  X  7)  cubic  units  in  the  back 
tier,  and  as  we  can  have  3  such  tiers,  the  number  of  cubic 
units  in  the  solid  will  be  represented  by  3  X  (5  X  7). 

Again,  if  we  form  a  pile  of  7  cubic  units  on  the  right-hand 
square  of  the  back  row,  then  another  pile  of  7  cubic  units  on 
the  next  square  in  front,  another  pile  of  7  cubic  units  on  the 
next  square  in  front,  we  shall  have  a  tier  of  (3  X  7)  cubic 
units.  We  can  have  5  such  tiers,  and  the  number  of  cubic 
units  in  the  solid  will  now  be  represented  by  5  X  (3  X  7). 

It  follows,  therefore,  that  the  total  number  of  cubic  units 
in  the  solid  may  be  represented  by 

7  X  (3  X  5),  or  by  3  X  (5  X  7),  or  by  5  X  (3  X  7). 

It  is  obvious  that  no  part  of  this  proof  depends  upon  the 
particular  numbers  3,  5,  and  7,  but  the  law  holds  for  any 
arithmetical  numbers  whatever,  and  may  be  expressed  by 

eX{aX  h)  =  a  X  (b  X  c)  =  b  X(aX  c). 

This  is  called  the  associative  law  of  addition  and  multi- 
plication, and  may  be  stated  as  follows: 

The  terms  of  an  expression,  or  the  factors  of  a  product, 
may  be  grouped  in  any  manner. 

30.   The  Index  Law. 

Since  a'  =  aa,  and  a^  =  aaa, 

a^Xa^  =  aaX  aaa  =  a^  =  a'*"*"' ; 
a*Xa  =  aaaa  X  a  =  a^  =  a^'*"^ 
If  a  stands  for  any  number,  and  m  and  n  for  any  integers, 

since       a^=  aaa to  m  factors, 

and         a**  =  aaa to  n  factors, 

a«  X  a''  =  (aaa to  m  factors).  X  (aaa to  «  factors), 

=  aaa to  (m-^n)  factors, 


16  SCHOOL    ALGEBRA. 

Hence,  the  index  law  may  be  stated  as  follows : 

The  index  of  the  product  of  two  powers  of  the  same  numher 
is  equal  to  the  sum  of  the  indices  of  the  factors. 

31.  These  four  laws,  the  commutative,  the  distributive, 
the  associative,  and  the  index  laws,  are  the  fundamental 
laws  of  Arithmetic,  and  together  with  the  law  of  signs, 
which  will  be  explained  hereafter,  they  constitute  the 
fundamental  laws  of  Algebra. 

32.  Quantities  Opposite  in  Kind.  If  a  man  gains  6  dollars 
and  then  loses  4  dollars,  his  actual  gain,  or,  as  we  com- 
monly say,  his  net  gain,  is  2  dollars  ;  that  is,  4  dollars'  loss 
cancels  4  dollars  of  the  6  dollars'  gain  and  leaves  2  dollars* 
gain.  If  he  gained  6  dollars  and  then  lost  6  dollars,  the  6 
dollars'  loss  cancels  the  6  dollars'  gain,  and  his  net  gain  is 
nothing.  If  he  gained  6  dollars  and  then  lost  9  dollars, 
the  6  dollars'  gain  cancels  6  dollars  of  the  9  dollars'  loss, 
and  his  net  loss  is  3  dollars.  In  other  words,  loss  and  gain 
are  quantities  so  related  that  one  cancels  the  other  wholly 
or  in  part. 

If  the  mercury  in  a  thermometer  rises  12  degrees  and 
then  falls  7  degrees,  the  fall  of  7  degrees  cancels  7  degrees 
of  the  rise,  and  the  net  rise  is  5  degrees.  If  it  rises  12  de- 
grees and  then  falls  12  degrees,  the  net  rise  is  nothing. 
If  it  rises  12  degrees  and  falls  15  degrees,  there  is  a  net 
fall  of  3  degrees.  In  other  words,  rise  and  fall  are  quan- 
tities so  related  that  one  cancels  the  other  wholly  or  in 
part. 

An  opposition  of  this  kind  also  exists  in  motion  forwards 
and  motion  backwards;  in  distances  measured  east  and 
distances  measured  west;  in  distances  measured  north  and 
distances  measured  south;  in  assets  and  debts;  in  time  be- 
fore and  time  after  a  fixed  date ;  and  so  on. 


INTRODUCTION.  17 

33.   Algebraic  Numbers.     If  we  wish  to  add  3  to  4,  we 
begin  at  4  in  the  natural  series  of  numbers, 

0123456789 


count  3  nnits  forwards,  and  arrive  at  7,  the  sum  sought.  If 
we  wish  to  subtract  3  from  7,  we  begin  at  7  in  the  natural 
series  of  numbers,  count  3  units  backwards,  and  arrive  at  4, 
the  difference  sought.  If  we  wish  to  subtract  7  from  7,  we 
begin  at ^7,  count  7  units  backwards,  and  arrive  at  0.  If 
we  wish  to  subtract  7  from  4,  we  cannot  do  it,  because 
when  we  have  counted  backwards  as  far  as  0  the  natural 
series  of  numbers  comes  to  an  end. 

In  order  to  subtract  a  greater  number  from  a  smaller  it 
is  necessary  to  assume  a  new  series  of  numbers,  beginning 
at  zero  and  extending  to  the  left  of  zero.  The  series  to  the 
left  of  zero  must  proceed  from  zero  by  the  repetitions  of  the 
unit,  precisely  like  the  natural  series  to  the  right  of  zero ; 
and  the  opposition  between  the  right-hand  series  and  the 
left-hand  series  must  be  clearly  marked.  This  oppesition 
is  indicated  by  calling  every  number  in  the  right-hand 
series  a  positive  number,  and  prefixing  to  it,  when  written, 
the  sign  -f ;  and  by  calling  every  number  in  the  left-hand 
series  a  negative  number,  and  prefixing  to  it  the  sign  — . 
The  two  series  of  numbers  will  be  written  thus : 


0    +1     +2    +3     +4 


Alqebkaic  Series  of  Numbers. 

If,  now,  we  wish  to  subtract  9  from  6,  we  begin  at  6  in 
the  positive  series,  count  9  units  in  the  negative  direction 
(to  the  left),  and  arrive  at  —  3  in  the  negative  series ;  that 
is,  6  -  9  =  -  3. 

The  result  obtained  by  subtracting  a  greater  number  from 
a  less,  when  both  are  positive,  is  always  a  negative  number. 


18  SCHOOL   ALGEBRA. 

In  general,  if  a  and  h  represent  any  two  numbers  of  the 
positive  series,  the  expression  a  —  h  will  be  a  positive  num- 
ber when  a  is  greater  than  h ;  will  be  zero  when  a  is  equal 
to  h ;  will  be  a  negative  number  when  a  is  less  than  h. 

In  counting  from  left  to  right  in  the  algebraic  series  num- 
bers increase  in  magnitude ;  in  counting  from  right  to  left 
numbers  decrease  in  magnitude.  Thus  —3,  —1,  0,  -|-2, 
+  4  are  arranged  in  ascending  order  of  magnitude. 

34.  We  may  illustrate  the  use  of  algebraic  numbers  as 
follows:  _5        0  8  20 

H \ H 


D 

Suppose  a  person  starting  at  A  walks  20  feet  to  the  right 
of  A,  and  then  returns  12  feet,  where  will  he  be?  Answer: 
At  (7,  a  point  8  feet  to  the  right  of  A  ;  that  is,  20  feet  — 12 
feet  =  8  feet ;  or,  20  -  12  =  8. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  20  feet,  w^here  will  he  be?  Answer:  At  A, 
the  point  from  which  he  started  ;  that  is,  20  —  20  =  0. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  25  feet,  where  will  he  now  be  ?  Answer :  At 
D,  a  point  5  feet  to  the  left  of  A  ;  that  is,  20  —  25  =  —  5 ; 
and  the  phrase  "  5  feet  to  the  left  of  A  "  is  now  expressed 
by  the  negative  quantity,  —  5  feet. 

35.  Every  algebraic  number,  as  -|-  4  or  —  4,  consists  of  a 
sign  -f  or  —  and  the  absolute  value  of  the  number.  The 
sign  shows  whether  the  number  belongs  to  the  positive  or 
negative  series  of  numbers;  the  absolute  value  shows  what 
place  the  number  has  in  the  positive  or  negative  series. 

When  no  sign  stands  before  a  number,  the  sign  -|-  is 
always  understood.  Thus  4  means  the  same  as  -|-  4,  a 
means  the  same  as  -f-  a.     But  the  sign  —  is  never  omitted. 


INTRODUCTION.  19 

36.  Two  algebraic  numbers  which  have,  one  the  sign  + 
and  the  other  the  sign  — ,  are  said  to  have  unlike  signs. 

Two  algebraic  numbers  which  have  the  same  absolute 
values,  but  unlike  signs,  always  cancel  each  other  when 
combined.     Thus  +4  —  4  =  0,  4-a  — a  =  0. 

37.  Double  Meanings  of  the  Signs  +  and  — .     The  use  of 

the  signs  -j-  and  —  to  indicate  addition  and  subtraction 
must  be  carefully  distinguished  from  the  use  of  the  signs  + 
and  —  to  indicate  in  which  series,  the  positive  or  the  nega- 
tive, a  given  number  belongs.  In  the  first  sense  they  are 
signs  of  operations,  and  are  common  to  Arithmetic  and 
Algebra ;  in  the  second  sense  they  are  signs  of  opposition, 
and  are  employed  in  Algebra  alone. 

38.  Addition  and  Subtraction  of  Algebraic  Numbers.     An 

algebraic  number  which  is  to  be  added  or  subtracted  is 
often  inclosed  in  a  parenthesis,  in  order  that  the  signs  + 
and  — ,  which  are  used  to  distinguish  positive  and  negative 
numbers,  may  not  be  confounded  with  the  +  and  —  signs 
that  denote  the  operations  of  addition  and  subtraction. 
Thus  +  4  +  (—  3)  expresses  the  sum,  and  -f  4  —  (—  3)  ex- 
presses the  difference,  of  the  numbers  +  4  and  —  3. 

In  order  to  add  two  algebraic  numbers  we  begin  at  the 
place  in  the  series  which  the  first  number  occupies  and 
count,  in  the  direction  indicated  hy  the  sign  of  the  second 
number,  as  many  units  as  there  are  in  the  absolute  value 
of  the  second  number. 

Thus  the  sum  of  -f  4  -f  (+  3)  is  found  by  counting  from 
-j- 4  three  units  in  the  positive  direction;  that  is,  to  the 
right,  and  is,  therefore,  +  7. 

The  sum  of  +  4  +  (—  3)  is  found  by  counting  from  +  4 
three  units  in  the  negative  direction ;  that  is,  to  the  left,  and 
is,  therefore,  -j- 1. 


20  SCHOOL   ALGEBRA. 

The  sum  of  —  4  +  (+  3)  is  found  by  counting  from  —  4 
three  units  in  the  positive  direction,  and  is,  therefore,  —  1. 

-5   -4   -3    -2   -1       0    +1    +2    +3    +4    +5    +6 


The  sum  of  —  4  -f  (—  3)  is  found  by  counting  from  —  4 
three  units  in  the  negative  direction,  and  is,  therefore,  —7, 

Hence,  to  add  two  or  more  algebraic  numbers,  we  have 
the  following  rules : 

Case  I.  When  the  numbers  have  lihe  signs.  Find  the 
sum  of  their  absolute  values,  and  prefix  the  common  sign  to 
the  result. 

Case  II.  When  there  are  two  numbers  with  unlike  signs. 
Find  the  difference  of  their  absolute  values,  and  prefix  to  the 
result  the  sign  of  the  greater  number. 

Case  III.  When  there  are  more  than  two  numbers  with 
unlike  signs.  Combine  the  first  two  numbers  and  this  result 
with  the  third  number,  and  so  on ;  or,  find  the  sum  of  the 
positive  numbers  and  the  sum  of  the  negative  numbers,  take 
the  difference  between  the  absolute  values  of  these  two  sums, 
and  prefix  to  the  result  the  sign  of  the  greater  sum. 

39.  The  result  is  called  the  sum.  It  is  often  called  the 
algebraic  sum,  to  distinguish  it  from  the  arithmetical  sum, 
that  is,  the  sum  of  the  absolute  values  of  the  numbers. 

40.  Subtraction.  In  order  to  subtract  one  algebraic  num- 
ber from  another,  we  begin  at  the  place  in  the  series  which 
the  minuend  occupies  and  count  in,  the  direction  opposite  to 
that  indicated  by  the  sign  of  the  subtrahend  as  many  units 
as  there  are  in  the  absolute  value  of  the  subtrahend. 

Thus,  the  result  of  subtracting  +  3  from  +  4  is  found  by 
counting  from  -f-  4  three  units  in  the  negative  direction ; 
that  is,  in  the  direction  opposite  to  that  indicated  by  the  sign 
+  before  3,  and  is,  therefore,  +1- 


INTRODUCTION.  21 

The  result  of  subtracting  —  3  from  +4  is  found  by  count- 
ing from  +  4  three  units  in  the  positive  direction ;  that  is, 
in  the  direction  opposite  to  that  indicated  by  the  sign  —  he- 
fore  3,  and  is,  therefore,  -f  7. 

The  result  of  subtracting  -f-3  from  —  4  is  found  by  count- 
ing from  —  4  three  units  in  the  negative  direction,  and  is, 
therefore,  —  7. 

The  result  of  subtracting  — 3  from  —4  is  found  by  count- 
ing from  —4  three  units  in  the  positive  direction,  and  is, 
therefore,  —  1. 

Collecting  the  results  obtained  in  addition  and  subtrac- 
tion, we  have 

Addition.  Subteactiof.    '. 

+44-(-3)=+4-3=+l.  -f4-(+3)=4-4-3=+l. 

+4+(+3)=+4  +  3=+7.  +4-(-3)=+4+3-+7. 

-,4-f(-3)=-4-3=-7.  -4-(+3)=-4-3=:-7. 

-44-(+3)=-4+3=-l.  _4-(-3)=-4+3=-l. 

No  part  of  this  proof  depends  upon  the  particular  num- 
bers 4  and  3,  and  hence  we  may  employ  the  general  symbols 
a  and  h  to  represent  the  absolute  values  of  any  two  alge- 
braic numbers.     We  shall  then  have 


Addition. 

Subtraction. 

+a+(-J)=+a-^>. 

+a-(+5)=+o-J. 

(1) 

-f-a+(+5)=+a+5. 

+a-(-J)=+a+J. 

(2) 

-a-f(-Z>)=-a-5. 

-a— (+S)=-a— 5. 

(3) 

-«+(-|-^)=-a+^». 

-a-(-J)=-a4-5. 

(4) 

From  (1)  and  (3),  it  is  seen  that  subtracting  a  positive 
number  is  equivalent  to  adding  an  equal  negative  number. 

From  (2)  and  (4),  it  is  seen  that  subtracting  a  negative 
number  is  equivalent  to  adding  an  equal  positive  number. 


22  SCHOOL    ALGEBRA. 

To  subtract  one  algebraic  number  from  another,  we  have, 
therefore,  the  following  rule  : 

Change  the  sign  of  the  subtrahend,  and  add  the  subtra- 
hend to  the  minuend. 

This  rule  is  consistent  with  the  definition  of  subtraction 
given  in  §  23 ;  for,  if  we  have  to  subtract  —  4  from  -f  3,  we 
must  add  +  4  to  the  subtrahend  —  4  to  cancel  it,  and  then 
add  -|-3  to  obtain  the  minuend;  that  is,  we  must  add  -\-*l 
to  the  subtrahend  to  get  the  minuend,  but  -f  7  is  obtained 
by  changing  the  sign  of  the  subtrahend  —4,  making  it  +4, 
and  adding  it  to  +  8,  the  minuend. 

41.  The  commutative  law  of  addition  applies  to  algebraic 
numbers,  for +4 +(— 3)  =  — 3 +  (+4).  In  the  first  case 
we  begin  at  +  4  in  the  series,  count  three  units  to  the  left, 
and  arrive  at  -f-l ;  in  the  second  case  we  begin  at  —  3  in 
the  series,  count  four  units  to  the  right,  and  arrive  at  + 1- 

The  associative  law,  also,  of  addition  is  easily  seen  to 
apply  to  algebraic  numbers. 

42.  Multiplication  and  Division  of  Algebraic  Numbers.     By 

the  definition  of  multiplication,  §  24. 

Since  +3  =  +  l  +  l  +  l; 

.•.3x(+8)  =  +  8  +  8  +  8 
=  +  24, 
and  3  X  (-  8)  -  -  8  -  8  -  8 

=  -  24. 


Again, 

since 

- 

-3  = 

:_1_1- 

-1; 

• 

••(- 

-3): 

X8  = 

—  24, 

-8 

id 

(- 

3): 

x(- 

-8)  = 

-(-8)- 

-(- 

8)- 

-(- 

-8) 

:+8+8+8 

= 

:  +  24. 

INTRODUCTION.  23 

No  part  of  this  proof  depends  upon  the  particular  num- 
bers 3  and  8.  If  we  use  a  to  represent  the  absohite  value 
of  any  number,  and  b  to  represent  the  absolute  value  of 
any  other  number,  we  shall  have 

(+a)x(+5)-  +  a5.  (1) 

(+a)x(-^)  =  -a5.  (2) 

(-  a)  X  (+  ^)  =  -  ab.  (3) 

{~a)x(-b)  =  -\-ab.  (4) 

43.  Law  of  Signs  in  Multiplication.  From  these  four  cases 
it  follows  that,  in  finding  the  product  of  two  algebraic 
numbers, 

Lihe  signs  give  +,  and  unlike  signs  give  — . 


44.   Law  of  Signs  in  Division. 

Since(+a)x(+^)  =  +  aJ,  .-.  +a5-H(-f  a)=+6. 
Since  (+a)x(-5)  =  — a5,  .-.  —ab^{-{-a)  =  —  b. 
Since  (— a)x(+5)  =  — a5,  .'.  —ab~{—a)=-{-b. 
Since  (— a.)x(-^)  =+a5,    /.  -\-ab-^{—a)  =  —  b. 


That  is,  if  the  dividend  and  divisor  have  like  signs,  the 
quotient  has  the  sign  + ;  and  if  they  have  unlike  signs, 
the  quotient  has  the  sign  — .     Hence,  in  division, 

Like  signs  give  + ;  unlike  signs  give  — . 

45.  From  the  four  cases  of  multiplication  that  we  have 
given  in  §  42  it  will  be  seen  that  the  absolute  value  of 
each  product  is  independent  of  the  signs,  and  that  the  signs 
are  independent  of  the  order  of  the  factors.  Hence  the  com- 
mutative and  associative  laws  of  multiplication  hold  for  all 
algebraic  numbers. 


24 


SCHOOL   ALGEBRA. 


46.  The  distributive  law  also  holds ;  for,  if 

a  (h -{-  c)  =  ah -{-  ac, 
then  —  a  (b  -{-  c)  =  —  ah  —  ac, 

and  (b  -\- c)  (—  a)  =  h  (^— a) -{- c  (—  a). 

Therefore,  for  all  values  of  a,  h,  and  c, 
a(h  -\-  c)  —  ah-}-  ac. 

From  the  nature  of  division  the  distributive  law  which 
applies  to  multiplication  applies  also  to  division. 

47.  We  have  now  considered  the  fundamental  laws  of 
Algebra,  and  for  convenience  of  reference  we  formulate 
them  below : 


a-{-(h-{- c)  =  a'{-h-^c 
a-{-(h  —  c)  =  a-{-h~c 
a~(b-^c)  =  a  —  b  —  c 
a  —  (h  —  e)  =  a  —  h-\-c 

(+a)x(+5)-  +  a^> 
(+  a)  X  (-  5)  =  -  ah 
(-  a)x{^h)==-  ah 
(-a)x(~h)  =  +  ah 

The  commutative  law : 

Addition  a-\-h  =  h-\-a'\ 

Multiplication  ah  =  ha       J 

The  associative  law : 

Addition  a -}- (h -\- c)  =  {a -{- h) -}- c 

Multiplication  a(hc)  =  (ah)c = ahc 


(1) 


(2) 


}■• 


(3) 


(4) 


INTEODUCTION.  25 

The  distributive  law : 

Multiplication      a(b-\-c)  ~ab  -^  ac} 

Division  (i±f)  =  *  +  £  •     •     •     •     (5) 

a  a     a      } 

The  index  law : 

Multiplication        a'^X  a''=  oT'^'' (6) 

These  laws  are  true  for  all  values  of  the  letters,  but  in  (6) 
m  and  n  are  for  the  present  restricted  to  positive  integral 
values. 

48.  Value  of  an  Algebraic  Expression.  Every  algebraic  ex- 
pression stands  for  a  number ;  and  this  number,  obtained 
by  putting  for  the  several  letters  involved  the  numbers  for 
which  they  stand,  and  performing  the  operations  indicated 
by  the  signs,  is  called  (he  value  of  the  expression. 

In  finding  the  values  of  algebraic  expressions,  the  begin- 
ner must  be  careful  to  observe  what  operations  are  actually 
indicated.     Thus, 

4 a  means  a-\-a-{-a-\-a;  that  is,  4  X  a. 
a*  means  a  X  a  X  a  X  a. 
^ahc  means  the  square  root  of  the  product  of  a,  5,  and  c. 
■y/ahc  means  the  product  of  the  square  root  of  a  by  he. 

Note,  The  radical  sign  V  before  a  product,  without  a  vinculum 
or  a  parenthesis,  affects  only  the  symbol  immediately  following  it. 

Va  +  h  means  that  h  is  to  be  added  to  the  square  root  of  a. 
•y/a-^-b  means  that  b  is  to  be  added  to  a  and  the  square 
root  of  the  sum  taken. 

49.  In  finding  the  value  of  a  compound  expression  the 
operations  indicated  for  each  term  must  be  performed  before 
the  operation  indicated  by  the  sign  prefixed  to  the  term. 


26  SCHOOL    ALGEBRA. 

Indicated  divisions  should  be  written  in  the  fractional  form, 
and  the  sign  X  omitted  between  a  figure  and  a  letter,  or 
between  two  letters,  in  accordance  with  algebraic  usage. 
Thus,  (i  —  tf)  -^  2  X  c  +  26  should  be  written 

^-^^+26. 
2c 

Note.  The  line  between  the  numerator  and  denominator  of  the 
fractions  serves  for  a  vinculum,  and  renders  the  parenthesis  un- 
necessary. ^ 

If  5  =  4,  and  <?  =  —  4,  the  numerical  value  is 
|^^]  +  26  =  ^+26  =  -l  +  26  =  25. 

Exercise  2. 

Note.  "When  there  is  no  sign  expressed  between  single  symbols 
or  between  compound  expressions,  it  must  be  remembered  that  the 
sign  understood  is  the  sign  of  multiplication. 

If  a  =  1,  h  =  2,  and  c  —  S,  find  the  value  of 

1.  la  — bo.  5.    2a  —  h-{-c.  9.    ^^ahc. 

2.  ao-\-h,  6.    ah-\-ho  —  ac.          10.    -V.Qabc. 

3.  Aah  —  c.  7.    h^+a'+c^  11.    c^-h\ 

4.  6ab-b  —  c.  8.    2ab-bbc\  12.    -s/c^—aK 

13.  a-2(b-\-c).  16.    65  — 10i(7^12a  +  2c. 

14.  {a-\-by-\-2{c-a)\        17.    b  c -t~  {b  ~  a)  -  {b -\- a). 

15.  -y/Wbc-ib-c).  18.    -VW?. 

If  a  =  1,  5  =  2,  d?  =  3,  and  c?  =  0,  find  the  value  of 

19.  7a  — 5c?  +  6d  21.    ^ab  —  cd—d, 

20.  ac-\'b  —  d.  22.    2  a  — 5  + c. 


J  INTRODUCTION.  27 

23.  ah -\- bo  — ad.  21.    h-~c-\-d. 

24.  2ah^bbc.  28.    a  — 2(5  +  c). 

25.  V^abcd.  29.    25  (3  -  5c)-^(a  -  2(?). 


26.    -^J^ahcd.  30.    2(a  +  57. 

Exercise  3. 

Remot^e  the  parentheses  (§  20),  and  find  the  'algebraic 
sum  of 

1.  (5-2)  +  (3  +  l).  11.    23  +  (2-7-6). 

2.  (6 -2) -(2 -3).  12.   24 -(2- 7-5). 
.3.    (-3  +  4)-(2  +  5).        13.    ll+(l-4-3). 

4.  _6-(2-3-l).  14.  (15-3-5)-7. 

5.  2-(5-7  +  8).  15.  10-(5-2-l). 

6.  8-(7-5  +  4).  16.  -17 -(5 -10- 13). 

7.  10-(5-6-7).  17.  -5  +  (3  +  2  +  7). 

8.  2-(3-3  +  4).  18.  -(2-3  +  4)-l. 

9.  (5-10)  +  (3-2).  19.  -(10-8-12  +  10). 
10.  -7+(3  +  2-4).  20.  +(8-2-3-1). 

If  a  =  l,  5  =  2,  and  c  =  — 3,  find  the  value  of 

21.  a-\-h-\-c.  24.  a-(— 5)  +  tf. 

22.  a  — 5  +  c.  25.  a  —  {—}))  —  c. 

23.  a-h-c.  26.  (- a)  +  (- 5)  +  (- c). 


+     - 


CHAPTER   II. 

ADDITION   AND    SUBTRACTION. 

Integral  Expressions. 

50.  If  an  algebraic  expression  contains  only  ^integral 
forms,  that  is,  contains  no  letter  in  the  denominator  of 
any  of  its  terms,  it  is  called  an  integral  expression.  Thus, 
3^ -\-l ca^  —  c^  —  k) G^ X,  ^ax—\hcy,  are  integral  expressions, 

but     ^  ~ — —c—x  .g  ^  fractional  expression. 

An  integral  expression  may  have  for  some  values  of  the 
letters  a  fractional  value,  and  a  fractional  expression  an 
integral  value..  If,  for  instance,  a  stands  for  j  and  h  for 
\,  the  integral  expression  2a  — 55  stands  for  f  —  f  =  J; 

and   the  fractional    expression  — -  stands  for  ^  -^  f  =  5. 

Integral  and  fractional  expressions,  therefore,  are  so  named 
on  account  of  the /or??i  of  the  expressions,  and  with  no  refer- 
ence whatever  to  the  numerical  value  of  the  expressions 
when  definite  numbers  are  put  in  place  of  the  letters. 

51.  A  term  may  consist  of  a  single  symbol,  as  a,  or  may 
be  the  product  of  two  or  more  factors,  as  6  a,  ab,  ba^bc.  If 
one  of  the  factors  is  an  arithmetical  symbol,  as  the  factor  5 
in  ba^bc,  this  factor  is  usually  written  first,  and  is  called 
the  coefficient  of  the  term ;  the  other  factors  are  called  literal 
factors. 

Note.  By  way  of  distinction,  a  factor  expressed  by  an  arithmeti- 
cal figure  is  called  a  numerical  factor,  and  a  factor  expressed  by  a 
letter  is  called  a  literal  factor. 


ADDITION   AND    SUBTRACTION.  29 

62.  Like  Terms.  Terms  which  have  the  same  combina- 
tion of  literal  factors  are  called  like  or  similar  terms;  terms 
which  do  not  have  the  same  combination  of  literal  factors 
are  called  unlike  or  dissimilar  terms.  Thus,  ba^bc,  —la^bc, 
a^hc,  are  like  terms,  but  ba^bc,  bab'^c,  babc^,  are  unlike 
terms. 

53.  A  simple  expression,  that  is,  an  expression  of  one 
term,  is  called  a  monomial.  A  compound  expression,  that 
is,  an  expression  which  contains  two  or  more  terms,  is 
called  a  polynomial.  A  polynomial  which  contains  two 
terms  is  called  a  binomial,  and  a  polynomial  which  contains 
three  terms  is  called  a  trinomial. 

54.  A  polynomial  is  said  to  be  arranged  according  to 
the  powers  of  some  letter  when  the  exponents  of  that  letter 
either  descend  or  ascend  in  the  order  of  magnitude.  Thus, 
^ao[^  —  ^bx'^—%ax-\-^b  is  arranged  according  to  the  de- 
scending powers  of  x,  and  8  6  —  6  aa;  —  4  5a;'' +  3  aa;^  is 
arranged  according  to  the  ascending  powers  of  x. 

55.  Addition  of  Integral  Expressions.  The  addition  of  two 
algebraic  expressions  can  be  represented  by  connecting  the 
second  expression  with  the  first  by  the  sign  +.  If  there 
are  no  like  terms  in  the  two  expressions,  the  operation  is 
algebraically  complete  when  the  two  expressions  are  thus 
connected. 

If,  for  example,  it  is  required  to  add  7n-\-n  —p  to 
a-\-b-\-  c,  the  result  will  he  a -\- b -\- c -{- {m -{- n  —  p)  \  or, 
removing  the  parenthesis  (§  20),  a -\- b -\- c -\- m, -\- n  —p. 

56.  If,  however,  there  are  like  terms  in  the  expressions  to 
be  added,  the  like  terms  can  be  collected;  that  is,  every 
set  of  like  terms  can  be  replaced  by  a  single  term  with  a 
coefficient  equal  to  the  algebraic  sum  of  the  coefficients  of 
the  like  terms. 


30  SCHOOL   ALGEBEA. 

If  it  is  required  to  add  5a^+4a  +  3  to  2a^  — 3a  — 4, 
the  result  will  be 

2a'^-3a-4  +  (5aH4a  +  3) 

=  2a'^-3a-4  +  5a'4-4a  +  3  §20 

=  2a^+5a='-3a  +  4a-4  +  3  §26 

This  process  is  more  conveniently  represented  by  arrang- 
ing the  terms  in  columns,  so  that  like  terms  shall  stand  in 
the  same  column,  as  follows  : 

2a''-3a-4 
5a'+4a  +  3 

la"-}-    a -I 

The  coefficient  of  a^  in  the  result  will  be  5  +  2,  or  7 ;  the 
coefficient  of  a  will  be  —  3  +  4,  or  1 ;  and  the  last  term  is 
-4  +  3,  or  —  1. 

Note.  "When  the  coefficient  of  a  term  is  1,  it  is  not  written,  hut 
understood ;  conversely,  when  the  coefficient  of  a  term  is  not  writ- 
ten, 1  is  understood  for  its  coefficient. 

If  we  are  to  find  the  sum  of  2a^  —  Sa^b  -{-  4:ab^  -{-  b^, 
a'  +  ^a'b-  7ab'  -  2b\  -  Sa'  +  a'b  ~  Sab'  -  45',  and 
2  a'  +  2  a'^^  +  6  ab'  —  3  b^,  we  write  them  in    columns. 

2a'-3a'b  +  4:ab'+    b' 

a'  +  4.a'b-7ab'-2P 

-Sa'+    a'b-drab'-U^ 

2a'  +  2a'b  +  6ab'-Sb' 

2a'-{-4:a'b  -Sb' 

The  coefficient  of  a^  in  the  result  will  be  2+1  —  3  +  2, 
or+2;  the  coefficient  of  a'^^  will  be  —3  +  4  +  1  +  2,  or  +4; 
the  coefficient  of  ab^  will  be  4  —  7  —  3  +  6,  or  0,  and, 
therefore,  the  term  ab^  will  not  appear  in  the  result ;  and 
the  coefficient  of  b^  will  be  1—2—4—3,  or  —8. 


ADDITION   AND   SUBTRACTION.  31 


Exercise  4. 

Add 

1.  7a,  2a,  —3a,  and  —5a. 

2.  7a;y,  2a:y,  —  4a;2/,  and  —bxy. 

3.  4a='^>,  -3a'5,  and  -Sa'^i. 

4.  Zxy,  ^xy,  lax,  and  —  Saa;. 

5.  a-{-h  and  a  —  h. 

6.  a:^  —  ^  and  a;^  — :r^ 

7.  5a;' +  6a; -2  and  3^^-70:  + 2. 

8.  3a;'-2a;y  +  y'  and  a;'-2a;y  +  3y^ 

9.  aa;'  +  ^»a;-4,  3 aa;^ -  2 ^>a;  +  4,  and  —  4aa;' -25a;  +  5. 

10.  5a;  +  3y  +  z,  3a;  +  2y  +  32;,  and  a;  — 3^- 52. 

11.  3a5  — 2aa;'+3a'a;,  4  a6  —  6  a'a;  +  5  aa;^  ah-\-a^x  —  aoi^, 

and  ax^  —  8  a5  —  5  a^x. 

12.  a*-2a'  +  3a'^-a+7,  2a*- 3a' +  2a''-a4-6,  and 

-a*-2a'  +  2a='-5. 

13.  ?>a^  -  ah  -^  ac  -2>h'' -\-4:bc  —  c\  -ba^-ab-ac-\-f)hc, 

-Uc-\-bc''  +  2ab,  and -4a''  + 5^-55^  + 2c\ 

14.  a;*  -  3a;' +  2a;' -4a; +7,  3a;*  + 2a;'  +  ^' —  5a;- 6, 

-4a;*  +  3a;'-3a;'  +  9^-2,  and  2a;*-a;'  +  a;'-a;+l. 

15.  3a;y'^-4a;'y  +  a;',  5a;'-llar2/'-12a;z',  — 7y'+a;V-^z', 

and  —  4  a;2;'  +  y'  —  z'. 

16.  a*-2a'  +  3a',  a^ -\- a^  ■\- a,  4a*+5a',  2a'  +  3a-2, 
^      and  -a' -2a  — 3. 

17.  a;'  +  2a;y'-a;V  — y',  2a;' -  3a;V  — 43;^*— 7/, 

and  a;'  — 8a;y'—  7y^. 


32  SCHOOL    ALGEBRA. 

57.  Subtraction  of  Integral  Expressions.  The  subtraction 
of  one  expression  from  another,  if  none  of  the  terms  are 
alike,  can  be  represented  only  hj  connecting  the  subtra- 
hend with  the  minuend  by  means  of  the  sign  — . 

If,  for  example,  it  is  required  to  subtract  a-\-b-\-c  from 
m-{-n  — p,  the  result  will  be  represented  by 

m-{-n  —p  —  {a -\- h -{- c)  ] 

or,  removing  the  parenthesis,  §  20, 

'm-\-n  —p  —  a  —  h  —  c. 

If,  however,  some  of  the  terms  in  the  two  expressions  are 
alike,  we  can  replace  two  like  terms  by  a  single  term. 

Thus,  suppose  it  is  required  to  subtract  a?—2a^-\-2a—l 
from  ^o?—2a^-{-a  —  2]  the  result  may  be  expressed  as 
follows : 

2,a^-2a'  +  a-2-  (a'  -  2a'  +  2a  -  1)  ; 

or,  removing  the  parenthesis '(§  20), 

3a'  -  2a'  +  a  -  2  ~-  a'  +  2a'  -  2a  +  1 
=  3a'-a'-2a'  +  2a'  +  a-2a-2  +  l 
=32a'-a-l. 

This  process  is  more  easily  performed  by  writing  the 
subtrahend  below  the  minuend,  mentally  changing  the  sign 
of  each  term  in  the  subtrahend,  and  adding  the  two  expres- 
sions.    Thus,  the  above  example  may  be  written 

3a'-2a'+    a  — 2 
a'-2a'  +  2a— 1 

2a'  -    a-1 

The  coefficient  of  a'  will  be  3  —  1,  or  2 ;  the  coefficient 
of  a'  will  be  —  2  -}-  2,  or  0,  and  therefore  the  term  a'  will 
not  appear  in  the  result ;  the  coefficient  of  a  will  be  1  —  2, 
or  —  1 ;  the  last  term  will  be  —  2  +  1,  or  —  1. 


ADDITION   AND    SUBTRACTION.  33 

Again,  suppose  it  is  required  to  subtract  a^+4aV~  3  aV 
—  4(2:r*  from  aV  +  2  aV  —  4  ao;*.  Here  terms  which  are 
alike  can  be  written  in  columns,  as  before  : 

aV  +  2aV  — 4a:i;* 
a^-f  4aV-3aV-4aa;* 


There  is  no, term  a^  in  the  minuend,  hence  the  coefficient 
of  a^  in  the  result  is  0—1,  or  —1;  the  coefficient  of  aV 
will  be  1  —  4,  or  —3;  the  coefficient  of  aV  will  be  2+3, 
or -r5  ;  the  coefficient  of  ax'^  will  be  —4  +  4,  or  0,  and 
therefore  the  term  do;*  will  not  appear  in  the  result. 

Exercise  5. 

1.  From  8a  — 45 -2c  take  2a -35 -3(7. 

2.  From  3a-45  +  3ctake  2a-85-c  — <^. 

3.  From  7a'  — 9:r  —  l  take  5a'  — 6a;- 3. 

4.  From  2a;'  —  2aa;  +  a'  take  x^  —  ax~  a'. 

5.  From  4a  — 35  — 3c  take  2a  — 35  + 4c. 

6.  From5a;'  +  7a;  +  4  take  3a;'-7a;  +  2. 

7.  From  2 aa;  +  3  5y  +  5  take  3 aa;  —  3 5y  —  5. 

8.  From4a'-6a5  +  25nake  3a'  +  a5  +  5'. 

9.  From4a'5  +  7a5'  +  9  take8-3a5'. 

10.  From  5  a^c  +  6  a'5  -  8 a'  take  5'  +  6  a'5  -  5  a'c. 

11.  From  a'  —  5'  take  ¥.  13,  From  5'  take  a?  —  h\ 

12.  From  a'  -  5'  take  a'.  14.  From  a'  take  a'  -  h\ 

15.    Froma;*  +  3aa;'  — 25a:'  +  3ca;  — 4c? 
take  3a;*+aa;'-45'  +  6ca;  +  c?. 


34  SCHOOL   ALGEBRA. 

If  A  =  Sa''-2ab  +  6h\  C  =  7a'-8ab-{-6b\ 

B^da'-bab-i-Sb',  I)  =  lla'-Sab-  46^ 
find  the  expression  for 

16.  A-\-C+B  +  I).  Id.    A  +  C-B-B. 

17.  A-O-Bi-B.  20.   A  ~a+B  +  B. 

18.  C-A-B+B.  21.   A+C-B  +  B. 

58.  Parentheses.  From  the  laws  of  parentheses  (§  20),  we 
have  the  following  equivalent  expressions  : 

a  +  (5  +  c)  =  a  +  5 -f  t?,  .'.  a-\-b -{- c  — a-{-(b-^c); 

a-\-(b~  c)  =  a-^b~  c,  .'.  a-}-b  —  c  =  a-^(b  —  c); 

a  —  (b -}-  c)  —  a  —  b  —  c,  .'.  a~b  —  c  —  a  —  (b -{- c); 

a  —  (b  ~  c)  =  a  —  b  -}-  c,  .'.  a  —  b-\-c  =  a  —  (b  —  c); 

that  is,  if  a  parenthesis  is  preceded  by  the  sign  +,  the 
parenthesis  may  be  removed  without  changing  the  sign 
of  any  term ;  conversely,  any  number  of  terms  may  be 
enclosed  within  a  parenthesis  preceded  by  the  sign  +, 
without  changing  the  sign  of  any  term. 
/  If  a  parenthesis  is  preceded  by  the  sign  — ,  the  paren- 
thesis may  be  removed,  provided  the  sign  of  every  terra 
within  the  parenthesis  is  changed,  namely,  -|-  to  —  and  —  to 
+  ;  conversely,  any  number  of  terms  may  be  enclosed 
within  a  parenthesis  preceded  by  the  sign  — ,  provided 
the  sign  of  every  term  enclosed  is  changed. 

59.  Expressions  may  occur  having  parentheses  within 
parentheses.  In  such  cases  parentheses  of  different  shapes 
are  used,  and  the  beginner,  when  he  meets  with  one 
branch  of  a  parenthesis  (,  or  bracket  [,  or  brace  {, 
must  look  carefully  for  the  other  part,  whatever  may  in- 
tervene ;  and  all  that  is  included  between  the  two  parts  of 


ADDITION   AND   SUBTRACTION.  35 

each  parenthesis  must  be  treated  as  the  sign  before  it  directs, 
without  regard  to  other  parentheses.  It  is  best  to  remove 
each  parenthesis  in  succession,  beginning  with  the  innermost 
one.     Thus, 

(1)  a-[h-{c-d)  +  e'\ 
=  a —\h  —  c -\-  d -\- e"] 
=  a  —  h-\-c  —  d~e. 

(2)  a-\h-[c-{d-e)-^n\ 
=  a~\h-[c-d+e+f]\ 
=^a  —  \b  —  c-{-  d—e—fl 

=  a  —  b-{-c  —  d-{-e  +/. 

Exercise  6. 

Simplify  the  following  by  removing  the  parentheses  and 
collecting  like  terms : 

1.  a  —  b  —  [a  —  (b  —  c)  —  c]. 

2.  m  ~  [n  —  (p  —  m)']. 

3.  2x-y  +  [Az-(y+2x)]l 

4.  Sa-l2b-[5c-(Sa  +  b)]]. 

5.  a-lb  +  [c-(d~b)  +  a]~2b\. 

6.  3a; -[9 -(2a; +  7)  + 3a;]. 

7.  2a; -[y- (a; -22/)]. 

8.  a -[25  + (3c -25)  + a]. 

9.  (a-a;  +  y)-(5-a;-y)  +  (a  +  6-2y). 

10.  3a -[-45  + (4a -5) -(2a -55)]. 

11.  4c-[a-(25-3c)  +  c]  +  [a-(25-5c-a)]. 

12.  a;  +  (y~z)-[(3a;-2y)  +  z]  +  [a;-(y-2z)]. 

13.  a-[2a+(a-2a)  +  2a]-5a-f6a  — [(a+2a)-a]S. 


36  SCHOOL   ALGEBRA. 

14.  2x~(Sy  +  z)-{b-(c-b)  +  c-[a~(c-b)]\. 

15.  a-[b-}-c~a  —  (a.i-b)~c]-i-(2a~'F+~c). 

Note.  The  sign  —  which  is  written  in  the  above  problem  before 
the  first  term  b  under  the  vinculum  is  really  the  sign  of  the  vinculum, 
—  b  +  c  meaning  the  same  as  —  (6  +  c). 


16.    10-x-l-x  —  [x-(x-5-x)]l 


17.  2x-  \2x-j-(y-z)-^zi-[2x-(y-z-27/)-Sz]+4:yl 

18.  a~[b-l-ci-a~(a-b)~c\]  +  [2a~(b-  a)  J. 

19.  a^  \b  ~[a  —  {c  —  b')  -\-c  —  a~{a  —  b  —  c)-~a\-\-a\. 

20.  ba-\-Za-[2>a  —  {2a~a-b)  —  a\-\-a\.  . 

Exercise  7. 

In  each  of  the  following  expressions  enclose  the  last  three 
terms  in  a  parenthesis  preceded  by  the  sign  — ,  remember- 
ing that  the  sign  of  each  term  enclosed  must  be  changed. 

1.  2a-b-2,cffd^^e^bf.\ 

2.  X  —  a  —  y  —  0  —  z  —  c. 

3.  a  +  5  — c  +  4a-5  +  l. 

4.  ax -\- by  -{- cz -\- bx  —  cy  -\-  cz. 

5.  3a  +  25  +  2c-5c?-3e  — 4/ 

6.  x  —  y-{-z  —  bxy  —  ^xz-{-2>yz. 

Considering  all  the  factors  that  precede  x,  y,  and  z,  respectively  as 
the  coefficients  of  these  letters,  we  may  collect  in  parentheses  the 
coefficients  of  x,  y,  and  z  in  the  following  expression  : 

ax  —  })y-\-ay  —  az  —  cz-{-hx  =  {a-\-'b)x-\-{a  —  h)y  —  {a-^  c)z. 

In  like  manner,  collect  the  coefficients  of  x,  y,  and  z  in 
the  following  expressions : 

7.  ax -{- by -\-  cz -\- bx  —  cy  -{■  az. 

8.  ax-{-2ay-\-^az  —  bx-{-?>y  —  ^bz  —  2z. 


ADDITION   AND   SUBTRACTION.  87 

9.  ax  —  2h^--bcz  —  4:bx-}-Sci/  —  7az.    - 

10.  ax  -\-Sa'i/-\-2b2/  —  bz  —  llcx-{-2cy  —  cz. 

11.  4:b^  —  Sax  —  6cz  -f  2bx  —  7 ex  —  bey  —  ex  —  ey—  ez. 

12.  6az~  bby-\-Scz  —  2bz  —  Say-{-bz  —  ax-^by. 

13.  z  —  by-\-Saz  —  Sey-\-2ax~2mx  —  bbz. 

14.  x-]-ay  —  az  —  acx  +  bcz  —  mny  —  y  —  z. 

Exercise  8. 

Examples  for  Review. 

1.  Add4:x^-5a''-bax^  +  6a%  Qa'i-Bx'-\-4:ax^+2a'x, 

Idax'-llx'  —  Wa'x,  and  10x'+7 a''x-\-5a'-lSax\ 

2.  Add  3ab-\-Sa  +  6b,  -ab-\-2a  +  4:b,   7ab~4:a-8b, 

and  6a  +  12b  —2ab. 

Note.  Similar  compound  expressions  are  added  in  precisely  the 
same  way  as  simple  expressions,  by  finding  the  sum  of  their  coeffi- 
cients.    Thus,  3. (a? r- y)  -\-  5{x  —  y)  —  2{x  —  y)  =  6{x  —  y). 

3.  Add4(5-a;),  6(5-a;),  3(5-.t),  and -2(5-^). 

4.  Add  (a -\-b)x''-\-{b  +  e)y^  +  (a  +  c)z\  (b  +  c)o(^ 

+  («  +  c)2/^  +  (a  +  5)z^  and  (a  +  e)x'  +  (a  +  ^»)3/' 
+  (^'  +  ^)^^ 
/5.    Kdd{ar\-h)x-\-{b^c)y^{e-\-a)z,-  {b -{-cyz+^c+o^x 
—  (p,-fb)y,  and  (arfc?)y+(«  +  ^)2^  — (^  +  ^)^• 
6.    From  a^  —  x"^  take  a?  '\-2ax-\-  x^. 

7.  From  ?>  a?  -]- 2  ax -{•  x^  take  a'  —  aa:  —  x"^. 

8.  From  8  a;^  —  3  ao;  +  5  take  5  a:' -f  2  ao; -f  5. 

9.  From  a^  -\-2>  Ve  -f  a^^  —  abc  take  a5^  —  abc  +  i'. 

10.  From  {a  -\-b^  X  -\-  {a  -\-  c)y  take  (a  —  5)  a;  —  (a  —  c)2/. 

11.  Simplify  7a-{3a  -  [4a-(5a- 2a)]5. 


38  SCHOOL   ALGEBRA. 

12.  Simplify  Sa-\a^b-[a-{-bi-c-(a+h-{- c  +  d)]]. 

13.  Bracket  the  coefBcients,  and  arrange  according  to  the 

descending  powers  of  x 

x^ —ax— c^x^  ~hx  -\-boi^  —  cx^  -\-  aV  —  a?— ex. 

14.  Simplify  a^-Q)'-c'-)-\h''-{G^-a^)'\-\-\&-{h''-a^)\ 

15.  If  a=l,  5  =  3,  c=5,  and  c?-=7,  find  the  value  of 

16.  From  2(^+lla+10^>-5f?  take  2^  +  5a-3&.    Find 

the  value  of  each  of  these  expressions  when  a,  6,  <?, 
and  d  have  the  values  1,  3,  5,  7,  respectively,  and 
show  that  the  difference  of  these  values  is  equal  to 
the  value  of  their  difference. 

17.  If  <2=1,  5  =-3,  c  =  -5,  c^=0,  find  the  value  of 

If  a  =  3,  5  =  4,  c  =  9,  and    2s  =  a  +  5+c,    find   the 
value  of 

18.  s(s  — a)(s  — 5)(s  — c). 

19.  s^  J^{8-  of  ^{s-hf  -\-{s-  c)\ 

20.  s' - (s  -  a) (s  -  5)  - (s  -  5)  (s  -c)-{8-c){8-a). 

21.  If  a;  =  «  +  25  — 3^,  y  =  5  +  2^-3a,  and  z  =  c  +  2a 

—  3  5,  show  that  ;r  -f-  y  -|-  2  =  0. 

22.  If  a:  =  a  — 25  +  3e,  y  =  5  — 2c  +  3a,  and  z  =  c  —  2a 

+  35,  show  thata;  +  y  +  z  =  2a  +  25  +  2c. 

23.  What  must  be  added  to  a;^  +  5y^  +  3 2^  in  order  that 

the  sum  may  be  2y^  —  z^l 

24.  What  must  be  added  to  5a^  —  7a^5  +  3«5^  in  order 

that  the  sum  may  be  a'  —  2  a^b  —  2  a^''  +  5^  ? 

25.  If  ^=5a^+3a^5-25^  F^Sa' --7 a'b -b\ 

G  =  2a'b~  a'--b\       H^  a'b  -  2a'  -  35', 
find  the  expression  for  E—  \_F—  (G  —  S^]. 


CHAPTER  III. 
MULTIPLICATION. 

Integral  Expressions. 

60.    The  laws  which  govern  the  operation  of  multiplica- 
tion are  formulated  as  follows  :  §  47 

ah  =  ha      ........     The  commutative  law. 


a  X  (be)  =  {ah)  X  c  =  abc 

a  (h -{- c)  ~  ah  -\-  ac  1 
a  (h~c)  =  ah  —  ac  J-^ 


a"*  X  a**  =  a" 


The  associative  law. 
The  distributive  law. 
The  index  law. 

The  law  of  signs. 


aX(-\-b)  =-\-ah 
a  X  (—  h)  ^~ah 
(-~q)xh  =  —  ah 
(-  a)x(-h)  =  i-ah  } 

61.  Multiplication  of  Monomials.  "When  the  factors  are 
single  letters,  the  product  is  represented  by  simply  writing 
the  letters  without  any  sign  between  them.  Thus,  the  prod- 
uct of  a,  h,  and  c  is  expressed  by  abc. 

62.  The  product  of  4a,  55,  and  3  c  is 

iaX  ^bxSc  =  4:XbxSahc  =  Q0abc. 

Note.  We  cannot  write  453  for  4  x  5  x  3  because  another  mean- 
ing has  been  assigned  in  Arithmetic  to  453,  namely,  400  +  60  +  3. 
Hence,  between  arithmetical  factors  the  sign  X  must  be  written. 


40  SCHOOL   ALGEBRA. 

63.  The  product  of  o^h  and  d^lP'  is 

64.  To  multiply  one  monomial  by  another,  therefore, 

Find  the  product  of  the  coefficients,  and  to  this  product 
annex  the  letters,  giving  to  each  letter  in  the  product  an  index 
equal  to  the  sum  of  its  indices  in  the  factors. 

Note.  The  beginner  should  determine  first  the  sign  of  the  product 
by  the  law  of  signs,  and  write  it  down ;  secondly,  after  the  sign  he 
should  write  the  product  of  the  coefficients ;  and  lastly,  each  letter 
Vith  an  index  equal  to  the  sum  of  its  indices  in  the  factors. 

65.  We  may  have  an  index  affecting  an  expression  as 
well  as  an  index  of  a  single  letter.  Thus,  {abcy  means 
ahc  X  ahc,  which  equals  aabhcc,  or  a'JV.  In  like  manner, 
{ahcf.  =  a%'"c\     That  is, 

The  nth  power  of  the  product  of  several  factors  is  equal  to 
the  product  of  the  nth  powers  of  the  factors. 

66.  By  the  law  of  signs,  we  have 

(-  a)x{—h)  =  -\-  ah, 
and  (+  ah)  X  (—<?)  =  —  ahc, 

that  is,         (—  a)  X(~h)x(—c)^  —  ahc  ; 
and  (—  ahc)  X  (—  d)  =  -\-  ahcd, 

that  is,         (—  a)  X{~h)X  (-  c)x(—d)  =  +  ahcd. 
It  is  obvious,  therefore,  that 

The  product  of  an  even  number  of  negative  factors  will 
be  positive,  and  the  product  of  an  odd  number  of  negative 
factors  will  be  negative. 


MULTIPLICATION.  41 

67.   Polynomials  by  Monomials.     We  have  (§  47), 

a  (b -{- c)  —  ab -{■  ac. 
In  like  manner, 

a(b  —  c-^-d—  e)  =  ab  —  ac-{-  ad—  ae. 
To  multiply  a  polynomial  by  a  monomial,  therefore, 

Multiply  each  temn  of  the  polynomial  by  the  monomial, 
and  add  the  partial  products. 

Exercise  9. 
Find  the  product  of 

1.  7cand5Z>.  6.  7abB,ndSae. 

2.  3a;  and  8y.  7.  —  2aand7aVy. 

3.  3(2'  and  6a^  8.  -  Sa'b  and  —  Sab\ 

4.  3  a  and  2 al  9.  —  5m7ip' and  —  4mW^^ 

5.  2mn  and  3mV  10.  —  8a^  —25',  and— 3a5. 

11.  —  2  x'^y,  xy^,  and  —  3  x'^y. 

12.  -  Sx^y,  —2a^b,  and  —x^a'^b'^. 

13.  5a  +  35and2a'. 

14.  ab  —  be  and  ba^bc. 

15.  aJ  —  ac  —  5c  and  aS^. 

16.  6a'5-7a'5'canda'5V 

17.  a'  +  5'-c'anda^5c^ 

18.  5a2-35'  +  2c2and4a5V. 

19.  a5c— 3a'5c'and  — 2a5'c. 

20.  —  xyz^  +  ^yz  and  —  x^yz. 

21.  —  2  m'w^'  —  mwp'  and  —  m'w^. 


42  SCHOOL   ALGEBRA. 

22.  x  —  y  —  z  and  —  3  a^y'z^. 

23.  —^ x^  s.iidix''-{- 2y''  —  z. 

24.  3a;— 2y  — 4and  Sar'. 

68.  Polynomials  by  Polynomials.  If  we  have  m-\-n-\-p 
to  be  multiplied  by  a  +  6  -}-  c,  we  may  substitute  M  for  the 
multiplicand  m-]-n-\-p  (§  12).    Then 

(a  +  Z»  +  <?)  jr=  aif  +  Z> Jf  +  c Jf.  §  28 

If  now  we  substitute  for  M  its  value  m-\-n-\-p,  we  have 
a(jn-\-n-\-p)-]rh(j^-{-n-\-p)-\-c(r)i-\-n-]-p') 
=  am  -{•  an  -j-  ap  +  Z>?^i  +  ^n  +  *^^/>  +  era  +  ^^i  +  <?pt 

To  find  the  product  of  two  polynomials,  therefore, 
Multiply  eoery  term  of  the  muUiplicand  hy  each  term  of 
the  multiplier,  and  add  the  partial  products. 

69.  In  multiplying  polynomials,  it  is  a  convenient  ar- 
rangement to  write  the  multiplier  under  the  multiplicand, 
and  place  like  terms  of  the  partial  products  in  columns. 

(1)    Multiply  5a— 6^  by  3a— 46. 

5a  -   6  5 

3a  -   4  Z> 

ISa'^-lSa^* 

-2Qa5  +  24Z>^ 

15a^-38a6  +  2452 

We  multiply  5  a,  the  first  term  of  the  multiplicand,  by 
3  a,  the  first  term  of  the  multiplier,  and  obtain  15  a'^;  then 

—  6  6,  the  second  term  of  the  multiplicand,  by  3  a,  and  ob- 
tain —  18a5.     The  first  line   of  partial  products  is   15 a^ 

—  18a6.  In  multiplying  by  —4  6,  we  obtain  for  a  second 
line  of  partial  products  —  20a6  +  246^  which  is  put  one 
place  to  the  right,  so  that  the  like  terms  — 18  ah  and 


MULTIPLICATION.  43 

— 20a5  may  stand  in  the  same  column.  We  then  add  the 
coefficients  of  the  like  terms,  and  obtain  the  complete  prod- 
uct in  its  simplest  form. 

(2)  Multiply  4a;  +  3  +  5a;'-  Gr'  by  4 -  Ba^'-Sa:. 

Arrange  both  multiplicand  and  multiplier  according  to 
the  ascending  powers  of  x. 

3+   4a;+    5ar^—   6a;' 
4_    5ar_    e>a^ 


12  +  16a;  +  20a;' -24a;' 

-  15a;  -  20a;'  -  25a;'  +  30a;* 
-18a;'-24a;'-30a;^  +  36a;^ 

12+      a;- 18a;'- 73a;'  +36a;^ 

(3)  Multiply  1  +  2a;  +  a;*  -  3 a;'  by  ar*  -  2  -  2a;. 

Arrange  according  to  the  descending  powers  of  x. 

a;*-3a;'  +  2a;+l 
a;' -2a;  -2 


a;'-3a;5  +  2a;*+    a;' 

-2a;*  +6a;'-4a;'-2a7 

-2a;*  +6.'r'-4a;  — 2 

a;' -5  a;**  +7  a;' +  2  a;' -6a;  —  2 

(4)  Multiply  o? -{•  h^  -{-  c^  —  ah  —  be  —  ac  hy  a -\- h -\- c. 

Arrange  according  to  descending  powers  of  a, 

a?  —  ah  —  ac-\-   h^  —      he -\-    (? 
a  ■\-    h-\'     c 

a'  —  a^b  —  a^c  +  ai'  —    ahc  +  ac' 

+  a'6  -a6'-    ahc  -{- h' -  h'c  +  hc^ 

+  a^o  —    ahc  —  ac^         +  h^c  —  hc^-\-(? 

a*  —Zahc  +6»  -\-& 


i) 


44  SCHOOL   ALGEBRA. 

Note.  The  student  should  observe  that,  with  a  view  to  bringing 
like  terms  of  the  partial  products  in  columns,  the  terms  of  the  multi- 
plicand and  multiplier  are  arranged  in  the  same  order. 

70.  A  term  that  is  the  product  of  three  letters  is  said  to 
be  of  three  dimensions,  or  of  the  third  degree.  In  general, 
a  term  that  is  the  product  of  n  letters  is  said  to  be  of  n 
dimensions,  or  of  the  nth  degree.  Thus,  fiohc  is  of  three 
dimensions,  or  of  the  third  degree ;  2  a^^V,  that  is,  2  aabbcc^ 
is  of  six  dimensions,  or  of  the  sixth  degree. 

71.  The  degree  of  a  compound  algebraic  expression  is  the 
degree  of  that  term  of  the  expression  which  is  of  highest 
dimensions. 

72.  When  all  the  terms  of  a  compound  expression  are  of 
the  same  degree,  the  expression  is  said  to  be  homogeneous. 
Thus,  a^-{-^s^y-{-Zxy^-\-y^  is  a  homogeneous  expression, 
every  term  being  of  the  third  degree. 

73.  The  product  of  two  homogeneous  expressions  is  homo- 
geneous. For  the  different  terms  of  the  product  are  found 
by  multiplying  every  term  of  the  multiplicand  by  each  term 
of  the  multiplier ;  and  the  number  of  dimensions  of  each 
partial  product  is  the  sum  of  the  number  of  dimensions  of 
a  term  of  the  multiplicand  and  of  a  term  of  the  multiplier 
counted  together.  Thus,  in  multiplying  a^  -\- h^  -\-  c^  —  ah 
'~hc  —  achja-[-h-{-c.  Example  (4),  each  term  of  the  mul- 
tiplicand is  of  two  dimensions,  and  each  term  of  the  multi- 
plier is  of  one  dimension ;  we  therefore  have  each  term  of 
the  product  of  2  + 1,  that  is,  three  dimensions. 

This  fact  affords  an  important  test  of  the  accuracy  of  the 
work  of  multiplication  with  respect  to  the  literal  factors ; 
for,  if  any  term  in  the  product  is  of  a  degree  different  from 
the  degree  of  the  other  terms,  there  is  an  error  in  the  work 
of  finding  that  term. 


MULTIPLICATION.  45 

74.  Any  expression  that  is  not  homogeneous  can  be 
made  so  by  introducing  a  letter,  the  value  of  which  is 
unity.  Thus,  in  Example  (3),  the  expressions  can  be  writ- 
ten x*  —  S aV -\-2a^x-{-  a*  and  a^  —  2a^x  —  2 a'.  The  prod- 
uct will  then  be  a;^  -  5aV  + 7aV  +  2aV  -  6a'a;- 2a^ 
which  reduces  to  the  product  given  in  the  example,  by 
putting  1  for  a, 

76.  It  often  happens  in  algebraic  investigations  that 
there  is  one  letter  in  an  expression  of  more  importance 
than  the  rest,  and  this  is  therefore  called  the  leading  letter. 
In  such  cases  the  degree  of  the  expression  is  generally  called 
by  the  degree  of  the  leading  letter.  Thus,  aV-f^or-f-c  is  of 
the  second  degree  in  x. 

Exercise  10. 
Find  the  product  of 

1.  a;  +  10  and  a;  +  6.  12.  2x  —3  and  a;  +  3. 

2.  a; ~  2  and  x—Z.  13.  :r  —  7  and  2x  —  1. 

3.  a;  —  3  and  a;  4- 5-  !'*•  w  — wand  2m +  1. 

4.  a: -|- 3  and  ar  —  3.  15.  m  — aandw-|-a. 

5.  a;— 11  and  a;— 1.  16.  3 a; H- 7  and  2a; -3. 

6.  ~a7  +  2  and  —  a;  — 3.       17.  5a;— 2y  and  5a;  + 2y. 

7.  —  a; -2  and  a;  — 2.  18.  3a;  — 4y  and  2a;  + 3y. 

8.  —  a;  4-  4  and  a;  —  4.  19.  x^  +  'if  and  o?  —  y^ 

9.  -a;+7anda;-}- 7.  20.  2ar' +  33/^^  and  a;'' -f  3/1 

10.  a;— 7  and  a; +7.  21.  a;  +  y +  z  and  a;  — y +  2. 

11.  a;  — 3  and  2a; -f- 3.  22.  a;  +  2y  — z  and  a;— y+2z. 

23.  a;^  —  a;y  -j-  y^  and  x^  -\-xy-\-  y'. 

24.  m^  —  ran  +  r^  and  tn  -f-  n. 

25.  m'  -f  mn  +  r^  and  m  —  n. 


46  SCHOOL    ALGEBRA. 

26.  o?~Zab^h''  and  a^  —  2>ah  —  b\ 

27.  a'^-7a  +  2anda'-2a  +  3. 

28.  2a;'-3a;y  +  4y'and  3a;'  +  4a:y  — Syl 

29.  x'^  +  xT/-}-  y^  and  x^  ~xz~  zl 

30.  x^  ^if"  -\-  '^  —■  xy  ~  xz  ~  yz  and  a;  +  y  +  z. 

31.  4a'^-10a5  +  25a'^^>'and5^»  +  2a. 

32.  a;^  +  4y  andy +  4a;. 

33.  a;'  +  22;y  +  8andy'  +  2a:y-8. 

34.  a^  +  ^j'^  +  1  —  a5  -  a  —  6  and  a  +  1  +  6. 

35.  3a;^-22/^  +  5z'^and8a:^  +  22/^-3zl 

36.  aj'^  +  y^  +  2a:y  —  2a;  — 23/  — 1  and  a;  +  y  — 1. 
■37.  a"'  +  2a'"-^-3a'"-'— 1  anda  +  1. 

38.  a"  —  4 a**-^  +  5 a"-'  +  a**"^  and  a  -  1. 

39.  a*"+^  —  4  a'"  +  2  a^'*"^  -  a""'  and  2 a=^  -  a'  +  a. 

40.  a;**  —  3/"+'  and  x""  +  3/"+^ 

Exercise  11. 
Simplify  : 

1.  (a-\-h-\-c){a-\-h-c)-{2ah-c'). 

2.  {m-\-n)m  —  \J^m  —  ny  —  n(ji  —  my]. 

3.  [a(?-(a-5)(^»  +  c)]-5[^>-(a-c)]. 

4.  (a7-l)(a;-2)-3a;(a;  +  3)  +  2[(a;  +  2)(a;  +  l)-3]. 

5.  4(a-3&)(a  +  35)-2(a~65)^-2(a^  +  65^). 

6.  (a;  +  y+  z)'-^(y+2;-a;)-y(a;  +  2;-y)-z(a;4-y-z). 

7.  b[{a  —  h)x—cy']  —  2[a{x  —  y)—hx'] 

~[^ax-{pc  —  2a)y\ 


MULTIPLICATION.  47 

Exercise  12. 

Examples  for  Review. 
Multiply 

1.  x^—x—19hjx^+2x—S. 

2.  l-\-2xi-x^  hj  l-a^-i-2x^^Sx. 

3.  2x^+2-{-Sx  hy  2x-Sx'-{-2a^, 

4.  Sx''-}-6  —  4:x  hj  S-\-6x'-7x. 

5.  x^-{-x—i/  by  x^  —  y^ -\- xy . 

6.  3  +  7a;'-5a;  by  8a;'^— 6a;-10a;'+4 

7.  5^+6a6'^-4a25  by  2a^h  ~  a}? -^a\ 

8.  a:^  +  aa;  — Z»  by  rt^+Sa?  — 4. 

9.  s^  —  mx^ ■\- nx  ■\- r  \)j  2^ -\- ex -{-  d. 

10.  x^  —  {a  -\-l>)  X  -\-  ah  by  a;  —  c?. 

11.  aj'  +  aj^^  +  a^y'  +  Z  by  y-a;. 

12.  4a;'  +  9y'— 6a:y  by  4a;'  + 9y'  + 6a;y. 

13.  a;*-3^'  +  5  by  a;'  +  4. 

14.  X*' —  x^y"^ -\- y^  by  aj^  +  ^y  +  y*. 

15.  2a;^-3a;'  +  4a;'-5  by  a;'-8. 

16.  0^'^  —  a'"y'"  +  3/^"*  by  a"*  +  3/"*. 

17.  a'^'-a'^^  +  a^'-l  by  a=«+l. 

18.  a^+5'+c'+2a5-ac  — ^'c?  by  a  +  ^>  +  c. 

19.  Simplify  (a  -  25)  (5  -  2a)  -  (a  -  3  ^')  (46  -  a)  +  2ah. 

20.  If  a  =  0,  6  =  1,  and  c  =  —  1,  find  the  value  of 

(a-6)(a  — c)  +  c(3a-5-c)  +  2ac-(a-c)  +  26. 

21.  [(2ar  +  y)H  (^ -  2y)^]  [(3a;  -  2yJ  -^  (2x  ~  3y)^. 

22.  a\b-c)~b^(a-c)  +  c'(a-b)-(a-h)(a-c)(b~c). 

23.  (2a-6)^+26(a+^)-3a'-(a-6)'4-(a+5)(a-5). 


CHAPTER  IV. 

DIVISION. 

Integeal  Expressions. 

76.    The  laws  for  division  are  expressed  in  symbols,  as 
follows  : 

+  a5  4-(+a)=  +  5 


-ah- 

-(+ 

a)  = 

■-~b 

-\-ah- 

-(- 

a)  = 

■-h 

-ab- 

-(- 

a)  = 

:  +  5 

a 

a     a 

• 

Law  of  signs. 


§44 


Distributive  law.    §  47,  (5) 


77.  The  dividend  contains  all  the  factors  of  the  divisor 
and  of  the  quotient,  and  therefore  the  quotient  contains  the 
factors  of  the  dividend  that  are  not  found  in  the  divisor. 

mv         ahc  aahx  12^ahc         o^  ^ 

Thus,    — =a;     — —-— ax  \    — - — -=  — oic. 
be  ab  —  4a6 

78.  If  we  have  to  divide  a^  by  a\  a^  by  a\  a'  by  a,  we 
write  them  as  follows  : 


aaaaa 


aa 


=  aaa  =  a^  =  a^~* ; 


a°  _  aaaaaa 
a*       aaaa 


aa 


a"     aaaa 


DIVISION.  49 

79.  To  divide  one  monomial  by  another,  therefore,  we  have 
the  following  rule  : 

Divide  the  coefficient  of  the  dividend  hy  the  coefficient  of 
the  divisor  {observing  the  law  of  signs),  and  subtract  the 
index  of  any  letter  in  the  divisor  from  the  index  of  that  letter 
in  the  dividend. 

Thus,   ^^2.,;    i|f  =7.5;    ^  =  -9^,.; 

Note.  Since  —  =  1,  and  also  by  the  rule  above  given,  —  =  a"-" 

=  o®,  it  follows  that  a*'  =  l.  Hence,  any  letter  which  by  the  rule 
would  appear  in  the  quotient  with  zero  for  an  index,  may  be  omitted 
without  affecting  the  quotient. 

80.  To  divide  a  polynomial  by  a  monomial,  we  have,  by  the 
distributive  law,  the  following  rule  : 

Divide  each  term  of  the  dividend  by  the  divisor,  and  add 
the  partial  quotients. 

mi         8a5-f  4ag  — 6ac?__8a^  .  ^ac     ^ad 
^^'  2a  ~  2a       2a       2a  " 

=  45  +  2c-3^. 

^a'b''x-12a'^bx''-?>a'x_^a'b''x    ^2a^bx''      ^d'x 
2>a^x  ^a^x         2>a^x        Sa'^x 

=  Sa'b'-4:abx-l. 


2a;'»-i  2x'' 


3^«+2_2a;~+i. 


Note.  Here  we  have  4n  +  l  — (2w  — l)  =  4n  + 1  — 2n  +  l  =  2n  +  2, 
and  3n  —  (2w  —  l)  =  3n  —  2w  +  l=»n  +  l,  as  indices  of  x  in  the  first 
and  last  terms  of  the  quotient  respectively. 


50  SCHOOL   ALGEBKA. 

Exercise  13. 
Divide 

1.  3a^  by  a^  13.  ahx^  by  ahx. 

2.  -42a;'by  6a;S.  14.  —2a^hx'\)j  oJ'x. 

3.  -35/by6z'.  15.  4:aho^hj~ax\ 

4.  -62"' by- 3  z.  16.  IWx'^hy  -^h'x\ 

5.  20a;y  by  -  ^.xy".  17.  256a;yz^  by  8:ryz. 

6.  -21a;*  by  -  1  x\  18.  -50a^5^  by  -lOa^'^^' 

7.  2^a*b^hj  -laP.  19.  845Vy^  by  14:ry^. 

8.  -25z^^>^by-5z^».  20.  -  30 a Vs  by  -  6 a;^. 

9.  —  24  mV  by  —  4  wVd^     21.  x^  -{-^xyhj  x. 

10.  —  35/^'-^  by  bpq\  22.    a'^  -  2a^'  by  a. 

11.  — 16rVby— 4rsl  23.    4a;'- Sar' by  2a;l 

12.  28m%"  by  4mV".  24.    —  6  a;'  -  2a;  by  -  2a;. 

25.  -8a^-16a'"by-8a^ 

26.  27 a' -36  a'' by  9a^ 

27.  -  30 a'  +  20 a'  by  -  10 a\ 

28.  -  12a;y  —  4a;y  by  -  4a;y. 

29 .  —  3  xh^  —  6  xh^  by  -  3  3^^. 

30.  3  a'b^c'  -  9  a^h'c'  by  3  a'5V. 

31.  x^  —  xy  —  xz  by  —  a;. 

32.  3a'-6a''5  — QaS^by  —  3a. 

33.  a;^3/'^  —  a;y  —  a;y  by  a;y. 

34.  (i'h'^c  —  o^V'c  —  d^bc^  by  ahc. 

35.  8a'-4a'^-6a52by-2a. 

36.  5  m^n  —  10 mV  —  15  mn^  by  5  mn. 


DIVISION.  51 

81.  To  divide  one  polynomial  by  another. 

If  the  divisor  (one  factor)         =  a-\-b  -{-  c, 

and  the  quotient  (other  factor)=  n-\-p-\-q, 

Ian-\-bn-\-  en 
■\- ap -\- hp -\- cp 
■^aq-{-hq-^  cq. 

The  first  term  of  the  dividend  is  an ;  that  is,  the  product 
of  a,  the  first  term  of  the  divisor,  by  n,  the  first  term  of  the 
quotient.  The  first  term  n  of  the  quotient  is  therefore 
found  by  dividing  an,  the  first  term  of  the  dividend,  by  a, 
the  first  term  of  the  divisor. 

If  the  partial  product  formed  by  multiplying  the  entire 
divisor  by  n  be  subtracted  from  the  dividend,  the  first  term 
of  the  remainder  ap  is  the  product  of  a,  the  first  term  of 
the  divisor,  by  p,  the  second  term  of  the  quotient ;  that  is, 
the  second  term  of  the  quotient  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  first  term  of  the  divisor. 
In  like  manner,  the  third  term  of  the  quotient  is  obtained 
by  dividing  the  first  term  of  the  new  remainder  by  the  first 
term  of  the  divisor ;  and  so  on.  Hence  we  have  the  fol- 
lowing rule : 

Arrange  both  the  dividend  and  divisor  in  ascending  or 
descending  powers  of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor. 

Write  the  result  as  the  first  term  of  the  quotient. 

Multiply  all  the  terms  of  the  divisor  by  the  first  term  of 
the  quotient. 

Subtract  the  product  from  the  dividend. 

If  there  be  a  remainder,  consider  it  as  a  new  dividend 
and  proceed  as  before. 


52 


SCHOOL   ALGEBRA. 


82.  It  is  of  fundamental  importance  to  arrange  tlie  divi- 
jdend  and  divisor  in  the  same  order  with  respect  to  a  com- 
mon letter,  and  to  keep  this  order  throughout  the  operation. 

The  beginner  should  study  carefully  the  processes  in  the 
following  examples : 

(1)  Divide  ^^  +  ISa;  +  77  by  07  +  7. 


a;^  +  18a;+77 
x'^    Ix 


x+1 


a;+ll 


lla;  +  77 
11:?;+ 77 


Note.  The  student  will  notice  that  by  this  process  we  have  in 
effect  separated  the  dividend  into  two  parts,  x^  -^-l  x  and  11a;  +  77, 
and  divided  each  part  by  a;  +  7,  and  that  the  complete  quotient  is 
the  sum  of  the  partial  quotients  x  and  11.     Thus, 

a;2  +  18a;  +  77  =  a;2  +  7a; -t- 11a;  +  77  =  (a;2  +  7a;)  +  (11a;  +  77); 
.  a;^+18a;  +  77.    a;^+7a;  ,  lla;  +  77_,,  .  n 

.  , » 1  -  3/  -f-  11. 

a;  +  7  a;  +  7  a;  +  7 

(2)   Divide  a^-2ah-\-h''  by  a -6.     ' 


a^-2ah  +  h' 
a^~    ah 


J  »-  <d^tA-t'-'<2-<n^ 


-  ah  +  h' 

-  ab  +  b' 


a  —  b     <f/'yf^'ii-€£*Z^ 


(3)    Divide  4 a V  —  4 aV -^x^—a^  by  x"" -  a\ 
Arrange  according  to  descending  powers  of  x. 


a;6_4aV+4aV- 


x^ 


a;*-3aV-i-a* 


-3aV+4aV-a« 
—  3(2V+3aV 


DIVISION. 


58 


(4)  Divide  22a^J»+ 156*  + 3a*- lOa'i  -  22a5» 
Arrange  according  to  descending  powers  of  a. 


3a*  -  lOa'J  +  22a'Z>''  -  22a5^  +  15 h' 
3a*-    e>a'h+    ^a^b^ 


-  ^a^b-\-l2>a?b^~22ab^ 

-  4:a'b-{.    8a'b'- 12  ab' 


a'-2ab  +  Sb^ 


Sa^~4:ab-^bb^ 


ba'b' 
&a'b' 


10  ab'+ 15  b* 
10a¥+15b' 


(5)  Divide  5r'-a:+l-3a;*byl  +  3a:''-^^ 
Arrange  according  to  ascending  powers  of  a;. 


1-    x+5a^ 
1  —  2^:4-3^'^ 


l-2a;  +  3a;'' 


3^* 

"l+    X  — 

x-Sx'^+bx^-Sx' 
x—2x'^^x' 

—  x''+2x'-Sx' 

—  x'4-2x'  —  Sx* 


(6)  Divide  x^ -}- y^  -{- z^  —  S  xyz  by  ar  +  y  +  z. 
Arrange  according  to  descending  powers  of  x. 


3?  -\-  yx^  +  zc^ 


x-{-y  +  z 


—  yx^ 


x"^  —  yx  —  zx -{•  y"^  —  yz -\-  z' 
zx^  —  Zyzx  +  3/^  +  z* 

H^ : 


y'^- 


zx^  +    y'^x  —  2yza;  +  2/^  +  z' 


~zx' 


yzx 


y'^x  —    yzx  +  z^a;  +  y*  +  z' 
^ +y'  +  y'z 


—    yzx  +  z'^x  —  y^z  + 


-  yz^ 


fz  -  yz" 


^x  +  y^  +  z* 
z^x  +  yz'  +  z' 


54  SCHOOL    ALGEBRA. 


.a!-6 


(7)  Divide  4  a*+'  —  30  a*  +  19  a*"*  +  5  a*-'  +  9  a' 
by  a'-'  —  7 a'-*  +  2a*-^  —Sa 

4  a'^+^-SO  a*+19  a"-^+  5  a*-'+9  a'-* 


a'-'-7a'~*+2a'-^-Sa'-^ 


^a'-2a'-3a' 


-  3 a*-i+21  a-'-^-e a^-'+9 «»=-* 

-  3  a'-^+21  a'-'-G  d'-'-\-9  a"-* 


Note,  "W^^ad  the  index  of  a  in  the  first  term  of  the  quotient  by- 
subtracting  ^^Kdex  of  a  in  the  first  term  of  the  divisor  from  the 
index  of  a  ii^^^  first  term  of  the  dividend.  Now  {x  +  V)  —  {x  —  3) 
='X  +  l—x  +  3  =  4:.  Hence  4  is  the  index  of  a  in  the  first  term  of 
the  quotient.    In  the  same  way  the  other  indices  are  found. 


Exercise  14. 
Divide 

1.  aH7«  +  12by  a  +  4.      6.  4a;'+ 12:p  +  9  by  2:^+3. 

2.  a'-^a  +  Qhy  a-B.       7.  G^r'^- lla;+4  by  3:r-4. 

3.  x'-]-2x^-}-fhjx  +  ^.    8.  8x''-10ax~Sa''hj4:X-}-a. 

4.  x'^-^2x7/-\-7/'^  hj  x  —  1/.     9.  3a^  — 4a  — 4  by. 2  — a. 

5.  x^~f\ij  x-y.  10.  a^-8a-3  by  3-a. 

11.  a*+ll«'— 12a-5aH6  by  3  +  a'-3a. 

12.  y^-9y^  +  y'-16y-4  by  2/^  +  4  +  43/. 

13.  36  +  m*-13m'  by  6  +  wH5w■ 
14.  l-s-3s''-s'  by  l  +  2s  +  s^ 

15.  h^-2h^-\-\  by  5^-25  +  1. 

16.  x''  +  2a;y  +  9/  by  a;^ -  2xy  +  3^^ 

17.  a^-\-h^\yj  a*'-a%-\-a%^~ah^-\-h''. 

18.  lH-5a;'-6^*  by  l  —  x^^^x". 


DIVISION.  55 

19.  8a;'3/H9y*+16a;*  by  4a;H32/''-4a^ 

^20.  a;^  +  2/'  +  z'+3a;V+3a;2/^  by  a;  +  2/  +  0. 

21.  a^-{-h^  +  c^  —  2>abc  by  a-\-h-\-c. 

22.  x^-\-^'i^-\-z^--^xyz  by  a;^  +  4?/'^H-z^  — ajz  — 2a:y  —  2y2;. 

23.  ^x^—^if-^-xy  —  xz  —  ^yz-z^  by  2a: +  3?/ +  2;. 

24.  x^  —  y'^—^yz  —  ^  by  a;4-2/  +  2- 

25.  a;*  +  a;^2/^  +  y*  by  x^ -{- xy -\r  y"^ . 

26.  a;*— 9a;'^  +  12:r-4  by  :^'+3a;-2. 

27.  y^-2y*-6/  +  42/H132/  +  6  by  yH  32/^^  +  3^  +  1- 

28.  y*-5yV  +  4z*  by  y'^-y2-22^ 

29.  a;'-4y2-92^+12yz  by  a;  +  2y-32. 

30.  a;^-41a;— 120  by  a;'+4rr  +  5. 

31.  a;*— 3  +  5:r  +  a7'-4a;'  by  3-2a;-a:l    ' 

32.  6-2a;*+10a;'-lla;Ha:  by  4a;-3-2a;^ 

'33.    l-6a;^+5:?;«  by  l-2a;  +  ^'. 

34.  a;*+81  +  9a;'  by  3:r-a;2_9^ 

35.  x^  —  'i/'  by  :r^  +  ^y-|-y^. 

36.  a;«  +  /  by  x' -\- y\ 

37.  a;*+aV  +  a*  by  a;?-aa;  +  a^ 

38.  a'-25^  +  a6-3c'+75c  +  2ac  by  3c  +  a  — 5. 

39.  a5  +  2a^-36'-45c-ac-(?^  by  c  +  2a  +  36. 

40.  15.«^-leTl*:c  +  4aV  +  ^«a:*--^«*  by  3a''+ Sooj-a;", 
/41.  a'— 86'-l-6a6  by  a-26-1. 

42.  a;''»-3a;'^"y^+3a;y"--y"»  by  a;"-y". 

43.  a*^"^"  -  4  a'^*'-^^^"  —  27  a"*+«-253n  _^  42  a"*+«-'5*" 

by  a"*  +  3  a'^-^^"  -  6  oT-'^b^'', 


66  SCHOOL   ALGEBRA. 

83.  Integral  expressions  may  hs^ve  fractional  coefficients, 
since  an  algebraic  expression  is  integral  if  it  has  no  letter  in 
the  denominator.  The  processes  with  fractional  coefficients 
are  precisely  the  same  as  with  integral  coefficients,  as  will 
be  seen  by  the  following  examples  worked  out : 

(1)  Add  ^a'-lab  +  {b\  and  f  a^-j-faS -|5^ 

^a'-^ab  +  ih' 

^a'^  +  ^ab-ib^ 

(2)  From  ia'-^ab  +  ib'  take  ^o^-^ab  +  ^bK 

:^a'-iab+   ^b' 
ia'-iab+   jb^ 

ia'-hiab-^b^ 

(3)  Multiply  ia'-iab  +  ^b'  by  ^a-^b. 

ia'~iab  +    ib^ 
ia-^b 


^a'-ia'b-i-    lab' 

-ia'b+   ^ab'-iP 


la'-id^b  +  ^ab'-ib' 

(4)     Divide    |^,3+1^5^2_44352^__5^^3    by    f5-|d 

ib'~mi'd-{--^-j^bd'-j\d^m  -jd 
ib'-   ^b'd  ib'-ibd-i-\d' 


^b'd-i--^i-bd'-^d' 
■f^b'd-h      bd' 

ibd'-^d' 
fbd'-^d' 


DIVISION.  67 

Exercise  15. 

1.  Add  la'b-hih'c'  +  -^  2ind- j\a'b-ib'c'-^, 

2.  From  fx'  +  Sax  —  la'  take  2x^  —  ^ax  —  ia', 

3.  From  iy  — fa  — |a;  +  ^^  take  i-y -f- J^a  — |a;. 

4.  mi\tiip\j  ic'-ic-ihj  ic'-ic  +  l 

5.  Uulti^lj  ^x-^x'i-lx'  hj  ^x  +  ^x'  +  ^x".      % 

6.  Multiply  0.5m*  — 0.4wV  +  1.2mV  +  0.8mn'— 1.4n* 

by  0.4  m' -0.6  WW  — 0.8  Til 

7.  Divide  ■5^a*-|a'5  +  i|a'^^'^  +  -l-a&3  ]3y  fa+i&.      .>^ 

8.  DiYide  -^d'+d'-^d'  +  id'  hj  -id''-\-2d.   y\ 

Exercise  16. 
'  Examples  foe  Revie-w", 

/i^  '^  ^'  -  ,  >-: , 

1.  Find  the  value  of  re'  +  y*  +  z'  ,^  3^%2;,  if  a;  =  1,  y  =  2, 

and  z  —  ~S. 

2.  Find  the  value  of  V25c  —  a,  and  of  ■\/2bc—a,  if  5  =  8, 

(?  =  9,  and  a  =  23. 

3.  Add  a=^5-a5''+5'  and  a'-|a'S  +  a5'-f6l    \ 

4.  Multiply  a'"'_a'"^>'"+5''"  by  a"*+^>'".        **' 

5 .  Multiply  4  «'"*+*  +  6  «*"+'  +9  a'  by  2  a"»+*  -  3  a\ 

6.  Divide  a;'+82/'—1250»+3Oa:yz  by  a;  +  2y- 50.    X 

7.  Simplify  (x -  of  -{x-by-(a-b)(a  +  b-  Sx). 

8.  Find  the  coefficient  of  a:  in  the  expression 

x-\-a  —  2[2a  —  b(c  —  x)]. 

9.  Multiply    4a;'«+2"-l_7a;2"-»«+2^5^2n+3m-2    ^y.   5^2-m-2«^ 


^ 


58  SCHOOL   ALGEBRA. 

^/lO.  Divide  cH^-''^  -  c'd^''''' -  c^'^d''-^  by  c'-'»c^^-^ 

J  11.  Divide  m ^  — '^^^'^V "^^  +  w^~*3/**~*  by  m^-y-*. 

(jL2.  Divide  a^+'^- a' +a'+^^  by  a^-'^+^J'. 

13.  Divide  a;  —  a:"*-^"+*  +  a;'"^  by  o;'-"*-'^'*. 

14.  Divide    ^  — y3-4m_p^4i,+l    ^^    ^2i_m+l^ 

15.  Divide  S^r^^**- 6:r'"2/'»  + 6a;y"- 2y*^"  by  a;"-2/«. 

16.  Divide  a?-1  ax?  +  a'o?  -  ahx  -  Z)'^;  +  a'^^*  +  a^" 

by  a;^  —  aa?  +  Z):i;  —  a5. 

17 .  Divide  a;*'"  +  a;'"*  +  1  by  a;""  —  a;"*  +  1.^'"" ' 

18.  Divide  3a;'^+^- 4a;'"+^- 12a;"*+^— 9a;*"+* 
by  a;'"+*-3a;"'+l 

19.  Divide  6a7*"*+^- ISaj^'^+H  13a;''"+^— 13a;'»+5- 5a;^ 
by  2a;'"*— 3a;'"— 1. 

20.  Divide  12a'^-'- a*"-'^- 20a^-^+ 19  a'"- 10a"+^ 
by  4a'"-3a"+'  +  2a'. 

21.  Arrange  according  to  descending  powers  of  a;  the  fol- 
lowing expression,  and  enclose  the  coefficient  of  each 
power  in  a  parenthesis  with  a  minus  sign  before  each 
parenthesis  except  the  first : 
a;'  —  2  ^a:  —  aV  —  ax  —  ax^  —ex  —  a^a^  —  hex. 

22.  Divide  1.2a*a;  — 5.494 aV-f  4.8 aV4- 0.9 aa;*—a;^ 
by  0.6aa7  — 2a;'. 

23.  Multiply  Ja*- -^a5  + ^5' by  ia+i^. 

24.  Multiply  ^a'+ab'i-ib'hj  ^a-^b. 

25.  ■DiYide^a^  +  ^ab^  +  ^jb^hj^a  +  ib. 

26.  Subtract  \a^  +  \xi/  +  -J-y'  from  ^a;'  —  |^a;y  -f  i3/*- 

27.  Subtract  a;'  +  -J^  a;3/  —  -J-y^  from  2  a;'  —  -J  a;y  +  y". 


CHAPTER  V. 

SIMPLE     EQUATIONS. 

84.  Equations.  An  equation  is  a  statement  in  symbols 
that  two  expressions  stand  for  the  same  number.  Thus, 
the  equation  3  a; +  2  =  8  states  that  Sx-\-2  and  8  stand  for 
the  same  number. 

85.  That  part  of  the  equation  which  precedes  the  sign 
of  equality  is  called  the  first  member,  or  left  side,  and  that 
which  follows  the  sign  of  equality  is  called  the  second  mem- 
ber, or  right  side. 

86.  The  statement  of  equality  between  two  algebraic 
expressions,  if  true  for  all  values  of  the  letters  involved, 
is  called  an  identical  equation ;  but  if  true  only  for  certain 
particular  values  of  the  letters  involved,  it  is  called  an 
equation  of  condition.  Thus,  (a  -}-  by  =  a^  -\-  2  ab  -{-  b^,  which 
is  true  for  all  values  of  a  and  b,  is  an  identical  equation; 
and  3.r-|-2  =  8,  which  is  true  only  when  x  stands  for  2,  is 
an  equation  of  condition. 

For  brevity,  an  identical  equation  is  called  an  identity, 
and  an  equation  of  condition  is  called  simply  an  equation. 

87.  We  often  employ  an  equation  to  discover  an  unknown 
number  from  its  relation  to  known  numbers.-  We  usually 
represent  the  unknown  number  by  one  of  the  last  letters  of 
the  alphabet,  as  a:,  y,  z  ;  and,  by  way  of  distinction,  we  use 
the  first  letters,  a,  b,  c,  etc.,  to  represent  numbers  that  are 
supposed  to  be  known,  though  not  expressed  in  the  number- 


60  SCHOOL   ALGEBRA. 

symbols  of  Arithmetic.  Thus,  in  the  equation  ax-\-h  =  c, 
X  is  supposed  to  represent  an  unknown  number,  and  a,  h, 
and  c  are  supposed  to  represent  known  numbers. 

88.  Simple  Equations.  An  integral  equation  which  con- 
tains the  first  power  of  the  symbol  for  the  unknown  number, 
X,  and  no  higher  power,  is  called  a  simple  equation,  or  an 
equation  of  the  first  degree.  Thus,  ax-\-h=^c  is  a  simple 
equation,  or  an  equation  of  the  first  degree  in  x. 

89.  Solution  of  an  Equation.  To  solve  an  equation  is  to 
find  the  unknown  number;  {hat  is,  the  number  which,  when 
substituted  for  its  symbol  in  the  given  equation,  renders  the 
equation  an  identity.  This  number  is  said  to  satisfy  the 
equation,  and  is  called  the  root  of  the  equation. 

90.  Axioms.  In  solving  an  equation,  we  make  use  of 
the  following  axioms : 

Ax.  1.  If  equal  numbers  be  added  to  equal  numbers, 
the  sums  will  be  equal. 

Ax.  2.  If  equal  numbers  be  subtracted  from  equal  num- 
bers, the  remainders  will  be  equal. 

Ax.  3.  If  equal  numbers  be  multiplied  by  equal  numbers, 
the  products  will  be  equal. 

Ax.  4.  If  equal  numbers  be  divided  by  equal  numbers, 
the  quotients  will  be  equal. 

*-  If,  therefore,  the  two  sides  of  an  equation  he  increased  hy, 

diminished  hy,  multiplied  by,  or  divided  hy  equal  numbers, 
the  results  will  be  equal. 

Thus,  if  8^  =  24,  then  8a7  +  4  =  24  +  4,  8a:-4  =  24-4, 
4x8a;  =  4x24,  and  8a; --4  =  24^4. 

^         91.   Transposition  of  Terms.     It  becomes  necessary  in  solv- 
ing an  equation  to  bring  all  the  terms  that  contain  the 


SIMPLE   EQUATIONS.  61 

symbol  for  the  unknown  number  to  one  side  of  the  equation, 
and  all  the  other  terms  to  the  other  side.  This  is  called 
transposing  the  terms.     We  will  illustrate  by  examples : 

(1)  Find  the  number  for  which  x  stands  when 

16a;- 11  =  7a: +70. 

The  first  object  to  be  attained  is  to  get  all  the  terms 
which  contain  x  on  the  left  side  of  the  equation,  and  all  the 
other  terms  on  the  right  side.  This  can  be  done  by  first 
subtracting  7^7  from  both  sides  (Ax.  2),  which  gives 

907-11  =  70,  ^ 

and  then  adding  11  to  these  equals  (Ax.  1),  which  gives 

9a;  =.81. 

If  these  equals  be  divided  by  9,  the  coefficient  of  x,  the 
quotients  will  be  equal  (Ax.  4)  ;  that  is,  a;  =  9. 

(2)  Find  the  number  for  which  x  stands  when  x-^b  =  a. 
The  equation  is  x-{-b  =  a. 

Subtract  b  from  each  side,  x-}-b  —  b  =  a  —  b.  (Ax.  2) 

Since  +  b  and  —  5  in  the  left  side  cancel  each  other 
(§  36),  we  have  x  =  a  —  b. 

(3)  Find  the  number  for  which  x  stands  when  x~  b  =  a. 
The  equation  is  x ~b  =^a. 

Add  +  £  to  each  side,  x —  b  -{-b  =  a-\-b.  (Ax.  1) 

Since  —h  and  +5  in  the  left  side  cancel  each  other 
(§  36),  we  have  x  =  a-\-b. 

(4)  What  number  does  x  stand  for  when  ax-\-b  =  cx-\-d1 

This  is  the  general  form  which  every  simple  equation  in 
x  will  assume  when  the  like  terms  on  each  side  have  been 


62  SCHOOL   ALGEBRA. 

collected.     In  this  equation  x  represents  the  unknown  num- 
ber, and  a,  5,  <?,  d  represent  known  numbers. 

If  now  we  subtract  (Ax.  2)  ex  and  h  from  each  side  of 
the  equation,  we  have 

ax  —  cx^d~h\ 

or,  bracketing  the  coefficients  of  x, 

{a  —  c)x=^  d—h. 

Whence,  dividing  both  sides  hj  a~c,  the  coefficient  of  x, 
we  get  ^_j. 


92.  The  effect  of  the  operation  in  the  preceding  equa- 
tions, when  Axioms  (1)  and*  (2)  are  used,  is  to  take  a  term 
from  one  side  and  to  put  it  on  the  other  side  with  its  sign 
changed.  We  can  proceed  in  a  like  manner  in  any  other 
case.     Hence  the  general  rule  : 

93.  Any  term  may  he  transposed  froTn  one  side  of  an  equa- 
tion to  the  other  provided  its  sign  is  changed. 

94.  Any  term,  therefore,  which  occurs  on  both  sides  with 
the  same  sign  may  be  removed  from  both  without  affecting 
the  equality. 

95.  The  sign  of  every  term  of  an  equation  may  be 
changed,  for  this  is  effected  by  multiplying  by  —  1,  which 
by  Ax.  3  does  not  destroy  the  equality. 

96.  Verification.  When  the  root  is  substituted  for  its 
symbol  in  the  given  equation,  and  the  equation  reduces  to 
an  identity,  the  root  is  said  to  be  verified.  We  will  illustrate 
with  examples : 

(1)   What  number  added  to  twice  itself  gives  24  ? 
Let  X  stand  for  the  number ; 


SIMPLE   EQUATIONS.  63 

then  2x  will  stand  for  twice  the  number, 

and  the  number  added  to  twice  itself  will  be  a;  -f  2a;. 

But  the  number  added  to  twice  itself  is  24 ; 
.'.x  +  2x  =  24:. 

Combining  x  and  2x,  3a;  =  24. 

Divide  by  8,  the  coefficient  of  a;,  x=  8.     (Ax.  4) 

The  required  number  is  8. 

Verification.  a;  +  2  a;  =  24, 

8  +  2  X  8  =  24, 

8  +  16  =  24, 

24  =  24. 

(2)  If  4  a;  — 5  stands  for  19,  for  what  number  does  x 
stand  ? 

We  have  the  equation  4  a;  —  5  =  19. 

Transpose  —  5  to  the  right  side,  4  a;  =  19  +  5. 

Combine,  4  a;  =  24. 
Divide  by  4,  x  —  6.  (Ax.  4) 

Verification.  4  a;  —  5  =  19, 

4  X  6  -  5  =  19, 

24  -  5  =  19, 

19  =  19. 

(3)  If  3a;— 7  stands  for  the  same  number  as  14  — 4a;, 
what  number  does  x  stand  for  ? 

"We  have  the  equation  3a;— 7  =  14  —  4a;. 

Transpose  —4  a;  to  the  left  side,  and 
—  7  to  the  right  side,  3.2;  +  4a;  =  14  +  7. 

Combine,  7a:  =  21. 

Divide  by  7,  a;=   3. 


64  SCHOOL   ALGEBRA. 

Verification.       Sa;— 7  =  14  —  4^, 

3x3-7  =  14-4x3, 
2  =  2. 

(4)    Solve  the  equation  (a;  —  3)  (a;  —  4)  =  a;  (a;  —  1)  —  30. 
We  have  (x-S)(x- 4:)  =  x(x-l)-S0. 

Remove  the  parentheses, 

x''-7x+12  =  x^~x-B0. 

Since  x^  on  the  left  and  x^  on  the  right  are  precisely  the 
same,  including  the  sign,  they  may  be  cancelled. 

Then  -  7a7+ 12  = -a;- 30. 

Transpose  —  a:  to  the  left  side,  and  +  12  to  the  right  side, 

-7a;  +  a;  =  -30-12. 
Combine, 

-6x  =  ~42. 
Divide  by  —  6,  x  =  7. 

Verification. 

(7 -3)  (7 -4)  =7(7-1) -30, 
4x3  =  7x6- 30, 
12  =  42-30, 
12  =  12. 

Exercise  17. 
Find  what  number  x  stands  for 

1.  If  a;  -  5  stands  for  7.  fe.  If  2x-6  =  7 -{-x. 

2.  If  a: +  8  stands  for  12.  J7.  If2a:-4  =  5  — a;. 

3.  If  6  a;- 12  stands  for  18.         -j  8.  If  5a;- 2  =  3a;  +  4. 

4.  If  7a;- 3  stands  for  25.        /  9.  If  7a;- 5  =  6a;- 1. 

5.  If  5a;+ 8  stands  for  43.        (lO.  If  5a;- 3  =  25 -2a;. 


SIMPLE   EQUATIONS.  65 

11.  If  3  a;  and  x-\~8  stand  for  the  same  number. 

12.  If  2x  —  5  stands  for  the  same  number  as  3a;. 
Solve  the  equations : 

13.  2a;-3  =  8  +  a;.  16.  3a;-4  =  12  — a?. 
TiT  5a;  4-  4  =  20  +  a;.  irT^a;  —  5  =  7  -  2a;. 
15.    2a;-3  =  7-a;.  18.    3a;+14  =  2-a;. 

Find  X 

19.  If  2a;  —  5  and  4a;  —  11  stand  for  the  same  number. 

20.  If  x(x  —  7)  and  x^~  70  stand  for  the  same  number. 

21.  If  a;(3a;  — 2)  and  3a;(a;  — 1)4-2  stand  for  the  same 

number. 

22.  If  3a;-5  =  4a;-10. 

23.  If  2a;  — 4  =  14 -a;. 

24.  If  3a;  —  8  and  4a;  —  11  stand  for  the  same  number. 

25.  If  2a;  —  5  and  7  —  a;  stand  for  the  same  number. 

26.  If  2a:' -23  and  (2a;  + 1)  (a;  — 3)  stand  for  the  same 

number. 

27.  If_(a;  +  3)(a;-7)-(a:-4)(a;+l)  =  25. 

28.  If  (2a; -1)  (a; +  3) -(a; -3)  (2a; -3)  =  72. 
Solve  the  equations : 

29.  a;(a;-5)  =  a;'— 30. 

30.  a;  (a; -I:  3)  =  a;' +18. 

31.  (a;-3)(a;+l)  =  a;'— 3a;  +  l. 

32.  (a;-l)(a;  +  2)  +  (a;  +  3)(a;-l)  =  2»(a;  +  4)-(a;  +  l). 

33.  a; (a;  +  8)  -  (a;  +  1)  (a;  -  2)  -  5  (a;  +  3)  4-  3  =  0. 

34.  (a;  -  3)  (a;  4-  3)  -  (a;  -  4)  (a;  4-  4)  -  a;  =  0. 


66  SCHOOL   ALGEBRA. 

97.  Statement  and  Solution  of  Problems.  The  difficulties 
which  the  beginner  usually  meets  in  stating  problems  will 
be  quickly  overcome  if  he  will  observe  the  following  direc- 
tions : 

Study  the  problem  until  you  clearly  understand  its  mean- 
ing and  just  what  is  required  to  be  found. 

Eemember  that  x  must  not  be  put  for  money,  length, 
time,  weight,  etc.,  but  for  the  required  number  of  specified 
U7iits  of  money,  length,  time,  weight,  etc. 

Express  each  statement  carefully  in  algebraic  language, 
and  write  out  in  full  just  what  each  expression  stands  for. 

Do  not  attempt  to  form  the  equation  until  all  the  state- 
ments are  made  in  symbols. 

We  will  illustrate  by  examples : 

(1)  John  has  three  times  as  many  oranges  as  James,  and 
they  together  have  32.     How  many  has  each  ? 

Let        X  stand  for  the  number  of  oranges  James  has ;     , 
then         3  a;  is  the  number  of  oranges  John  has  ; 
and   a;  +  3  a;  is  the  number  of  oranges  they  together  have. 

But  32  is  the  number  of  oranges  they  together  have ; 
.•.a;  +  3a;  =  32; 
or,  4  a;  =  32, 

and  a;  =  8. 

Since  a;  =  8,  3  a;  =  24. 

Therefore  James  has  8  oranges,  and  John  has  24  oranges. 
Note.   Beginners  in  stating  the  preceding  problem  generally  write : 
Let  X  ==  what  James  had. 

Now,  we  know  what  James  had.  He  had  oranges,  and  we  are  to 
discover  simply  the  number  of  oranges  he  had. 

(2)  James  and  John  together  have  $24,  and  James  has 
$8  more  than  John.     How  many  dollars  has  each? 


SIMPLE    EQUATIONS.  67 

Let  X  stand  for  the  number  of  dollars  John  has  ; 

then  a;  +  8  is  the  number  of  dollars  James  has ; 

and  a;  +  (a;  +  8)  is  the  number  of  dollars  they  both  have. 

But  24  is  the  number  of  dollars  they  both  have ; 

.-.  a;  +  (a;  +  8)  =  24. 

Removing  the  parenthesis,  we  have 

a:  +  a;  +  8  =  24. 
Transposing  and  collecting  like  terms,  we  have 

2a; -16. 
Dividing  by  2,  we  get 

a;  =  8. 
Since  a;  =  8,  a;  +  8  =  16. 

Therefore  John  has  $8,  and  James  has  $16. 

Note.   The  beginner  must  avoid  the  mistake  of  writing 

Let  X  =  John's  money.  ^ 

We  are  required  to  find  the  number  of  dollars  John  has,  and  there- 
fore X  must  represent  this  required  number. 

(3)  The  sum  of  two  numbers  is  18,  and  three  times  the 
greater  number  exceeds  four  times  the  less  by  5.  Find  the 
numbers. 

Let  X  =  the  greater  number. 

Then,  since  18  is  the  sum,  and  x  is  one  of  the  jiumbers,  the  other 
number  must  be  the  sum  minus  x.     Hence 

18  —  a;  =  the  smaller  number. 

Now,  three  times  the  greater  number  is  3  x,  and  four  times  the  less 
number  is  4  (18  —  x). 

We  know  from  the  problem  that  3  a;  exceeds  4(18  — a;)  by  5;  or, 
in  other  words,  we  know  that  the  excess  of  3  a;  over  4(18  —  x)  equals 
5.  It  only  remains  to  determine  what  sign  the  word  "excess"  im- 
plies. If  we  are  in  doubt  about  it,  we  can  apply  the  phrase  to  two 
arithmetical  numbers.  We  shall  have  no  difficulty  in  seeing  that  the 
excess  of  50  over  40  is  10;  that  is,  50  —  40,  and  hence  that  the  sign 
—  is  implied  by  the  word  "  excess." 


68  SCHOOL    ALGEBRA. 


Hence, 

3  a;  -  4(18 -x)  =  the  excess 

But 

5  =  the  excess. 

.•.3a; -4(18 -a)  =  5, 

3a; -72  + 4a;  =  5. 

.•.7a;  =  77. 

d 

a;  =  11. 

or 


and 

Therefore  the  numbers  are  11  and  7. 

(4)  Find  a  number  whose  treble  exceeds  40  by  as  much 
as  its  double  falls  short  of  35. 

Let  X  =  the  required  number ; 

then  3  a;  =  its  treble, 

and  3  a;  —  40  =  the  excess  of  its  treble  over  40 ; 

also,  35  —  2a;  ==  the  number  its  double  lacks  of  35 

Hence,  3a;-40  =  35  -  2a;. 

Transposing,  3  a;  +  2  a;  =  35  +  40, 
.•.5a;  =75, 

and  X  =  15. 

Therefore  the  number  required  is  15. 

(5)  Find  a  number  that  exceeds  50  by  10  more  than  it 
falls  short  of  80. 

Let  X  =  the  required  number ; 

then  a;  —  50  =  its  excess  over  50, 

and  80  —  a;  =  the  number  it  lacks  of  80. 

Hence,  a;  —  50  —  (80  —  a;)  =  the  excess. 

But  10  =  the  excess. 

...  a;  -  50  -  (80  -  a;)  =  10, 
or  a;  -  50  -  80  +  a;  =  10. 

.•.2a;  =140, 
and  X  =  70. 


Therefore  the  number  re(juired  is  70. 


SIMPLE  EQUATIONS.  69 

Exercise  18. 

1.  If  a  number  is  multiplied  by  7,  the  product  is  301. 
Find  the  number. 

2.  The  sum  of  two  numbers  is  48,  and  the  greater  is  five 
times  the  less.     Find  the  numbers. 

3.  The  sum  of  two  numbers  is  25,  and  seven  times  the 
less  exceeds  three  times  the  greater  by  35.  Find  the  num- 
bers. 

4.  Divide  20  in  two  parts  such  that  four  times  the 
greater  exceeds  three  times  the  less  by  17. 

5.  Divide  23  into  two  parts  such  that  the  sum  of  twice 
the  greater  part  and  three  times  the  less  part  is  57. 

Y  6.  Divide  19  into  two  parts  such  that  the  greater  part  ex-^ 
ceeds  twice  the  less  part  by  1  less  than  twice  the  less  part. 

*^  7.  A  tree  84  feet  high  was  broken  so  that  the  part 
broken  off  was  five  times  the  length  of  the  part  left  stand- 
ing.    Eequired  the  length  of  each  part. 

/'      8.    Four  times  the  smaller  of  two  numbers  is  three  times 
/    the  greater,  and  their  sum  is  63.     Find  the  numbers. 

L^  9.  A  farmer  sold  a  sheep,  a  cow,  and  a  horse  for  $216. 
He  sold  the  cow  for  seven  times  as  much  as  the  sheep,  and 
the  horse  for  four  times  as  much  as  the  cow.  How  much 
did  he  get  for  each  ? 

^^JLD.  Distribute  $15  among  Thomas,  Eichard,  and  Henry 
so  that  Thomas  and  Eichard  shall  each  have  twice  as  much 
as  Henry. 

y  11.  Three  men.  A,  B,  and  C,  pay  $1000  taxes.  B  pays 
four  times  as  much  as  A,  and  C  pays  as  much  as  A  and  B 
together.     How  much  does  each  pay  ? 


70  /  SCHOOL   ALGEBRA. 

12.  John's  age  is  three  times  the  age  of  James,  and 
their  ages  together  are  16  years.    What  is  the  age  of  each  ? 

13.  Twice  a  certain  number  increased  by  8  is  40.  Find 
the  number. 

14.  Three  times  a  certain  number  is  46  more  than  the 
number  itself.     Find  the  number. 

15.  One  number  is  four  times  as  large  as  another.  If  I 
take  the  smaller  from  12  and  the  greater  from  21,  the 
remainders  are  equal.     What  are  the  numbers? 

16.  The  joint  ages  of  a  father  and  son  are  70  years.  If 
the  age  of  the  son  were  doubled,  he  would  be  4  years 
younger  than  his  father.     What  is  the  age  of  each  ? 

17.  A  man  has  6  sons,  each  4  years  older  than  the  next 
younger.  The  eldest  is  three  times  as  old  as  the  youngest. 
What  is  the  age  of  each  ? 

l^^^  Add  $24  to  a  certain  amount,  and  the  sum  will  be 
a,s  much  above  $80  as  the  amount  is  below  $80.  What  is 
the  amount  ? 

19.  Thirty  yards  of  cloth  and  40  yards  of  silk  together 
cost  $330;  and  the  silk  costs  twice  as  much  per  yard  as 
the, cloth.     How  much  does  each  cost  per  yard? 

20.  Find  the  number  whose  double  diminished  by  24 
exceeds  80  by  as  much  as  the  number  itself  is  less  than  100. 

21.  In  a  company  of  180  persons  composed  of  men, 
women,  and  children  there  are  twice  as  many  men  as 
women,  and  three  times  as  many  women  as  children. 
How  many  are  there  of  each? 

22.  A  banker  was  asked  to  pay  $56  in  five-dollar  and 
two-dollar  bills  in  such  a  manner  as  to  pay  the  same  num- 
ber of  each  kind  of  bills.  How  many  bills  of  each  kind 
must  he  pay?  ,.j^  .'/, 


X 


,     X  SIMPLE   EQUATIONS.  71 

23 Allow  can  $3.60  be  paid  in  quarters  and  ten-cent 
pieces  so  as  to  pay  twice  as  many  ten-cent  pieces  as 
quarters?  '  '      ' 

24.    I  have  $  1>98  in  ten-cent  pieces  and  three-cent  pieces, 
and  have  four  times  as  many  three-cent  pieces  as  ten-cent 
^        pieces.x^  How  many _^vje.  I  of  eacR  ?  ^  ^  ^  ^^^^  /^2-^ 

Note,  In  problems  involving  quantities  of  the  same  kind  ex- 
pressed in  different  units,  we  must  be  careful  to  reduce  all  the  quanti- 
ties to  the  %ame  unit. 

^       /'   25.    I  have  $17  dollars  in  two-dollar  bills  and  twenty- 
;^  Tvfive-cent  pieces,  and  have   twice  as  many  bills  as  coins. 
^  How  many  have  I  of  each  ? 

(         26.    I  have  $6.50  in  silver  dollars  and  ten-cent  pieces, 
\^nd  I  have  20  coins  in  all.     How  many  have  I  of  each  ? 

V  27.  A  bought  9  dozen  oranges  for  $2.00.  For  a  part  he 
paid  20  cents  per  dozen ;  for  the  remainder  he  paid  25  cents 
a  dozen.     How  many  dozen  of  each  kind  did  he  buy  ? 

28.  A  gentleman  gave  some  children  10  cents  apiece, 
and  found  that  he  had  just  50  cents  left.  If  he  had  had 
another  half-dollar,  he  might  have  given  each  of  them  at 
first  20  cents  instead  of  10  cents.  How  many  children 
were  there? 

29.  A  is  twice  as  old  as  B  and  6  years  younger  than  C. 
The  sum  of  the  ages  of  A,  B,  and  C  is  96  years.  What  is 
the  age  of  B  ? 

30.  Divide  a  line  24  inches  long  into  two  parts  such 
that  the  one  part  shall  be  6  inches  longer  than  the  other. 

/  31.  Two  trains  travelling,  one  at  25  and  the  other  at  30 
miles  an  hour,  start  at  the  same  time  from  two  places  220 
miles  apart,  and  move  toward  each  other.  In  how  many 
hours  will  the  trains  meet  ? 


/A 


X 


72  SCHOOL   ALGEBRA. 

32.  A  man  bought  twelve  yards  of  velvet,  and  if  he  had 
bought  1  yard  less  for  the  same  money,  each  yard  would 
have  cost  $  1  more.     What  did  the  velvet  cost  a  yard  ? 

33.  A  and  B  have  together  $  8 ;  A  and  0,  $  10 ;  B  and  C, 
$  12.     How  much  has  each  ? 

34.  Twelve  persons  subscribed  for  a  new  boat,  but  two 
being  unable  to  pay,  each  of  the  others  had  to  pay  $4  more 
than  his  share.     Find  the  cost  of  the  boat. 

35.  A  man  was  hired  for  26  days  on  condition  that  for 
every  day  he  worked  he  was  to  receive  $3,  and  for  every  day 
he  was  idle  he  was  to  pay  $1  for  his  board.  At  the  end  of 
the  time  he  received  $58.     How  many  days  did  he  work? 

36.  A  man  walking  4  miles  an  hour  starts  2  hours  after 
another  person  who  walks  3  miles  an  hour.  How  many 
miles  must  the  first  man  walk  to  overtake  the  second  ? 

37.  A  man  swimming  in  a  river  which  runs  1  mile  an 
hour  finds  that  it  takes  him  three  times  as  long  to  swim  a 
mile  up  the  river  as  it  does  to  swim  the  same  distance  down. 
Find  his  rate  of,swimming  in  still  water. 

38.  At  an  election  there  were  two  candidates,  and  2644 
■    votes  were  cast.     The  successful  candidate  had  a  majority 

of  140.     How  many  votes  were  cast  for  each  ? 

39.  Two  persons  start  from  towns  55  miles  apart  and 
walk  toward  each  other.  One  walks  at  the  rate  of  4  miles 
an  hour,  but  stops  2  hours  on  the  way ;  the  other  walks  at 
the  rate  of  3  miles  an  hour.  How  many  miles  will  each 
have  travelled  when  they  meet  ? 

\jy  40.    A  had  twice  as  much  money  as  B  ;  but  if  A  gives  B 
^$10,  B  will  have  three  times  as  much  as  A.     How  much 

has  each? 

41.    If  2  a;  —  8  stands  for  20,  for  what  number  will  4  —  a; 

stand  ? 


SIMPLE   EQUATIONS.  73 

42.  A  vessel  containing  100  gallons  was  emptied  in  10 
minutes  by  two  pipes  running  one  at  a  time.  The  first  pipe 
discharged  14  gallons  a  minute,  and  the  second  9  gallons  a 
minute.     How  many  minutes  did  each  pipe  run  ? 

43.  A  man  has  8  hours  for  an  excursion.  How  far  can 
he  ride  out  in  a  carriage  which  goes  at  the  rate  of  9  miles 
an  hour  so  as  to  return  in  time,  walking  at  the  rate  of  3 
miles  an  hour  ?    . 

44.  If  Sx  —  4ia  =  2x  —  a,  find  the  number  for  which 
4:X~1a  stands. 

2x  —  3a 


45.    li  7 X  ~  a  =  9  (x  -—  a),  find  the  number 


x-{-a 


Note.  When  we  compare  the  ages  of  two  persons  at  a  given  time, 
and  also  a  number  of  years  after  or  before  the  given  time,  we  must 
remember  that  both  persons  will  be  so  many  years  older  or  younger. 
Thus,  if  a  man  is  now  2x  years  old  and  his  son  x  years  old,  5  years 
ago  the  father  was  2a;  —  5  and  the  son  x—5,  and  5  years  hence  the 
father  will  be  2a;  +  5  and  the  son  x  +  5,  years  old, 

46.  A  man  is  now  twice  as  old  as  his  son  ;  15  years  ago 
he  was  three  times  as  old  as  his  son.     Find  the  age  of  each. 

47.  A  man  was  four  times  as  old  as  his  son  7  years  ago, 
Ind  will  be  only  twice  as  old  as  his  son  7  years  hence.    Find 

/the  age  of  each. 

tl8.    A,  who  is  25  years  older  than  B,  is  5  years  more 
in  twice  as  old  as  B.     Find  the  age  of  each. 
19.    A. man  is  25  years  older  than  his  son  ;  10  years  ago 
was  six  times  as  old  as  his  son.     Find  the  age  of  each. 

50.  The  difference  in  the  squares  of  two  consecutive 
numbers  is  19.     Find  the  numbers. 

51.  The  difference  in  the  squares  of  two  successive  odd 
numbers  is  40.     Find  the  numbers. 


CHAPTER  VI. 
MULTIPLICATION  AND  DIVISION. 

Special  Rules. 

98.  Special  Eules  of  Multiplication.  Some  results  of  mul- 
tiplication are  of  so  great  utility  in  shortening  algebraic 
work  that  they  should  be  carefully  noticed  and  remem- 
bered.    The  following  are  important : 

99.  Square  of  the  Sum  of  Two  Numbers. 

=  a{a-\-h)-{-h(a-\-h) 
=--a^-]-ah-\-ah-\-h'' 
=  a'  +  2ab-]-h\ 

Since  a  and  b  stand  for  any  two  numbers,  we  have 

Rule  1.  The  square  of  the  sum  of  two  numbers  is  the 
sum  of  their  squares  plus  twice  their  product. 

100.  Square  of  the  Difference  of  Two  Numbers. 

{a-by  =  {a-b){a-b) 

=  a(a  —  b)  —  b  (a  —  b) 

=  a^  —  ab  —  ab-\-b^ 

=  a'  —  2ab  +  b\ 
Hence  we  have 

Rule  2.  The  square  of  the  difference  of  two  numbers  is 
the  sum  of  their  squares  minus  twice  their  product. 


SPECIAL   RULES   OF   MULTIPLICATION.  75 

101.  Product  of  the  Sum  and  Difference  of  Two  Numbers. 

(a  ■\-b){a—h)  =  a(a  —  h)-\-h{a  —  h) 
—  a^  —  ah -{- ab  —  h^ 

=  0?-  h\ 

Hence,  we  have 

Rule  3.    The  product  of  the  sum  and  difference  of  two 
numbers  is  the  difference  of  their  squares. 

If  we  put  2x  for  a  and  3  for  h,  we  have 
Rule  1,  {2x  +  3/  -  4a;^  +  12a;  +  9. 

Rule  2,  {2x  -  3)2  =  4a;^  -  12a;  +  9. 

Rule  3,  (2a;  H-  3)  (2a;  -  3)  =  4a;'^  -  9. 

Exercise  19. 
Write  the  product  of 

1.  {x^y)\  7.  {x^y){x-y) 

2.  {x-a)\  8.  (42;-3)(4z  +  3). 

3.  {x^-2h)\  9.  (3a'^  +  452)(3a''-46^). 

4.  (3a;-2c)^  10.  (3a  — c)(3a- c). 
6.    (4y-5)^  11.  (x-\-1b'')ix+1b% 

6.    {^a^-\-4:zy.  12.    {ax  +  2by){ax  —  2by). 

102.  If  we  are  required  to  multiply  a+Z>+cby  a-\-h^c^ 
we  may  abridge  the  ordinary  process  as  follows  : 

(jrjL-\-h-\-c){a-\-b-c)=^[{a-\-h)-\-c\[{a-\-h)-c] 

By  Rule  3,  =  (a  +  by  -  c» 

By  Rule  1,  =  a*  +  2a6  +  i» -- c». 


76  SCfiOOii   ALGEfeftA. 

If  we  are  required  to  multiply  a-{-b  —  c  by  a—h-\-c,  we 
may  put  the  expressions  in  the  following  forms,  and  per- 
form the  operation  : 

{a  +  b  -  c){a-h  -{-  c)  =  [a-\-  (h  -  c)][a-  {h  -  c)] 
By  Kule  3,  =  o?  -  (h  -  cj 

By  Rule  2,  =  a^  -  (h'  -2hc-{-  c') 

=  a'-b'  +  2bc-c\ 

Exercise  20. 
Find  the  product  of 

1.  x-{-^-\-z  and  x—y—z. 

2.  x  —  y-^z  and  x—y  —  z. 

3.  ax-\-by-\-l  and  ax-{-by—l. 

4.  \-\-x  —  y  and  l—x-\-y. 

5.  a  +  25  — 3c  and  a  — 25  +  3(?. 

6.  o^-ab  +  b''  2.Tidi  o?-{-ab  +  b\ 

7.  m^  +  m?2  -}-  n^  and  m^  —  mn  +  ?2^. 

8.  2  +  ^  +  ^^  and  2  —  x  —  x'^. 

9.  a^  +  (^  +  1  and  a^  —  a  +  1. 
10.    3a;+2y  — 2  and  3a;— 2y+2;. 

103.   Square  of  any  Polynomial.     If  we  put  x  for  a,  and 

y  -j-  2  for  b,  in  the  identity 

(a+by  =  a'i-2ab  +  b\ 
we  shall  have 

[x+(y-}-z)J  =  x^  +  2x(y  +  z)-{-(y  +  z)\ 
or         (a;  +  yH-z)'    =a;^  +  2a;y+2a;2;  +  y2  +  2yz  +  z» 
^x'  +  y^  +  z'-\-2xy+2xz-{-2yz. 


SPECIAL   RULES   OF   MULTIPLICATION.  77 

It  will  be  seen  that  the  complete  product  consists  of  the 
sum  of  the  squares  of  the  terms  of  the  given  expression  and 
twice  the  products  of  each  term  into  all  the  terms  that  fol- 
low it. 

Again,  if  we  put  a  —  h  for  a,  and  c—dior  h,  in  the  same 
identity,  we  shall  have 

[(a-h)  +  (o-d)J 

=  {a~by-j-2(^a-h){c-d)  +  (c-dy 
=  (a'-2abW)+2a(c-d)-2b(c-d)-{-(c'-2cd-i-d') 
=  d'-2ab-\-b'+2ao-2ad—2bci-2bdi-c'-2cd-\-d^ 
=  a''+b''i-c''+d'~2ab  +  2ao-2ad-2bc  +  2bd~2cd. 

Here  the  same  law  holds  as  before,  the  sign  of  each 
double  product  being  +  or  — ,  according  as  the  factors  com- 
posing it  have  like  or  unlike  signs.  The  same  is  true  for 
any  polynomial.     Hence  we  have  the  following  rule  : 

KuLE  4.  The  square  of  a  polynomial  is  the  sum  of  the 
squares  of  the  several  terms  and  twice  the  products  obtained 
by  multiplying  each  term  into  all  the  terms  that  follow  it. 

Exercise  21. 
"Write  the  square  of 

1.  2x-^y.  10.  aj'  +  y'  +  z*. 

2.  a-|-5  +  c.  11,  2x~y  —  z. 

3.  x-[-y  —  z.  12.  a~2b  —  ^c. 

4.  x-y-^z.  13.  2>a  —  b  +  2c. 

5.  a;  +  y  +  5.  14.  x-{-2y  —  ^z. 

6.  a;  +  2y  +  3.  15.  o^  +  y  +  z  +  l. 
1.    a-b-{-c.  16.  4n:  +  y-f2-2. 

8.  3x-2y-f4.  17.    2x  —  y  —  z-^. 

9.  2a;-3y  +  42.  18.   a;-2y  — 3z  +  4. 


78  SCHOOL    ALGEBRA. 


104.  Product  of  Two  Binomials.  The  product  of  two  bino- 
mials which  have  the  form  x-{-a,  x-{-b,  should  be  carefully 
noticed  and  remembered. 

(1)  (x+^){x  +  S)  =  x(x  +  S)  +  5(x-\-S) 

=  x^  +  Sx+6x  +  15 
=  x'  +  Sx  +  16. 

(2)  (x-b)(x-3)  =  x  (x-S)-6  (x-S) 

=  x''  —  Sx-bx  +  16 
=  a;' -8a; +  15. 

(3)  {x  +  6)(x-S)=x(x-S)  +  5(x-  3) 

=  x^-Sx-\-5x-lb 
=  x'  +  2x~15. 

(4)  (x-6){x  +  S)  =  x(x+S)-b(x  +  S) 

=  x''  +  Sx-bx-lb 
=  x'-2x-15. 

Each  of  these  results  has  three  terms. 

The  first  term  of  each  result  is  the  product  of  the  first 
terms  of  the  binomials. 

The  last  term  of  each  result  is  the  product  of  the  second 
terms  of  the  binomials. 

The  middle  term  of  each  result  has  for  a  coefficient  the 
algebraic  sum  of  the  second  terms  of  the  binomials. 

The  intermediate  step  given  above  may  be  omitted,  and 
the  products  written  at  once  by  inspection.     Thus, 

(1)  Multiply  a;  +  8  by  a;  +  7. 

8+7  =  15,    8x7  =  56. 
,\{x-\-S)(x+1)  =  x''+16x  +  6Q. 


SPECIAL    RULES    OF   MULTIPLICATION.  79 

(2)  Multiply  x-^hj  X--1. 

(_8)  +  (_7)  =-15,    (_8)(-7)  =  +  56. 
.-.  (a;  -  8)  (a;  -  7)  =  a;^  -  15ar  +  56. 

(3)  Multiply  x—lyhjx-\-  6y. 

-7+6  =  -l,     (-7y)x6y  =  -42y». 
.-.  (x - 7y) (ix  +  67/)  =  x'-xy- 422/^ 

(4)  Multiply  x^  +  6  (a  -{- b)  hj  x^  -  6  (a -\-  b). 

+  6-5-1,   6(a+b)x-5(a  +  b)=-B0(a-\-by. 
.-.  [x'+6(a-\-b)]  [x'-5(a-{-b)]  =  x*-j-{ai-b)x'-Z0(a+by. 

Exercise  22. 
Find  by  inspection  the  product  of 


1. 

(^  +  8)  (a; +  3). 

16. 

(^^-9)(a;^  +  8). 

2. 

(^  +  8)  (^-3).          • 

17. 

(x'+2y')(x'~Sf). 

3. 

(^_7)(^_|_10). 

18. 

(x^^8f)(x'-4.y'y    ^ 

4. 

(a; -9)  (^-5). 

19. 

(ab-8)(ab  +  b). 

5. 

(a; -10)  (a; +  9). 

20. 

(ab-1xy){ab-{-2>xy). 

6. 

(a -10)  (a -5). 

21. 

(x-Sy)(x-Sy). 

7. 

(x-Sa)(x  +  2a). 

22. 

(x+6)(x  +  &y 

8. 

(a-\-2b)(a-4:b). 

23. 

(a-Sb)(a~Sb). 

9. 

(a -12)  (a -3). 

24. 

(x-c){x-d). 

10. 

(a  +  2b)(a  +  ^b). 

25. 

(x  +  a)(x-b). 

11. 

(a-^b)(a  +  7b). 

26. 

(x-a)(xi-b). 

12. 

(a  +  2b)(a-9b). 

27. 

[(a  +  b)  +  2][(a+b)-^l 

13. 

{x-Sa)(x-4:a). 

28. 

[(^+y)-2][(rr+y)+4]. 

14. 

ix-{-^z)(x~2z). 

29. 

(x  +  y~7)(x  +  yi-10). 

15. 

(x  +  6y)(a:-5y). 

30. 

(x-y-1)(x-i/-10). 

80  SCHOOL  ALGEBRA. 

105.  In  like  manner  the  product  of  any  two  binomials 
may  be  written. 

(1)  Multiply  2a-^>  by  3«  +  4^>/ 

=  6a^  +  5a6-46^ 

(2)  Multiply  2a;  +  3y  by  Sa;  — 2y. 
The  middle  term  is 

i^x)  X  (-  2y)  +  3y  X  3:r  =  5a;y ; 
.-.  (2a;  +  Zy) (3 a;  -  2y)  =  6a;'  +  5a:y  -  6y^ 

Exercise  23. 
Find  the  product  of 

1.  2>x  —  y  and  2a; +  y.  6.  10a;  — 3y  and   10a;— 7y. 

2.  4a;-3y  and  3a;-2y.     7.  3a'-25'  and  2a'+3^>^ 

3.  5a;  — 4y  and  3a;  — 4y.     8.  a'  +  ^^  and  a  —  h. 

4.  a;-7y  and  2a;  — 5y.        9.  3a''— 25'  and  2a  +  35. 

5.  11a;  — 2y  and  7a;  +  y.    10.  o^—lP-  and  a +  6. 

106.  Special  Eules  of  Division.  Some  results  in  division 
are  so  important  in  abridging  algebraic  work  that  they 
should  be  carefully  noticed  and  remembered. 


a-\-h.    Hence 


Rule  1.    The  difference  of  the  squares  of  two  nwtnbers  is 
divisible  by  the  sum,  and  by  the  difference,  of  the  numbers. 


107.   Difference  of  Two  Squares. 

From  §101,(a  +  ^>)(a^ 

~b)  =  a'- 

-b\ 

.    a'- 5' 

. .   ~  =Cb  - 

a  +  b 

—  b,  and 

a'- 
a  — 

-b' 
-b 

SPECIAL    RULES    OF    DIVISION.  81 

Exercise  24. 

Write  by  inspection  the  quotient  of 


1. ^-  14. 

a- 2                        .  ab'c'  +  x^ 

2     ^~^'  15     ^•'^'  -  ^'" 

3.    1^-^'  16.    ^^-(^  +  ^)' 

4  + a  '     a  — (6  +  c) 

:r-5'  *     a-(36-4c)* 

5.    ^^^.  18.    l-(^-y)l 

6  4- a;  l  +  (ar  — y) 

Sa-b'  '      (3a;  — y)  +  4z' 

^     25a;^-365'  2^     (a;  +  3  a)' -  9  o:^ 

5a;  +  6^>    '  '     (X'i-3a)-Sx' 

^9c'-d'  l~(7a-5by 

'      Ic-d  '  '    l  +  (7a-6b)' 

9     9^^-l  22     (3^+2y?-4.^ 

3a+l  '    (3a;  +  2y)-2z 

'     4  — 2a  '     (a-b)-{-(c--y) 

^^     9a' -2W  24     ^'-(y  +  ^)' 

3a'^  +  5y^  *     ^-(y  +  2) 

^2     ^x''-9y'  25     (^-2yy-25 

2:r«-3y^  *     (a;-2y)  +  5 


13. 


4:r^°-a«         .  2g     {2x-\-yy~9z' 

23^ -a*'  '    (2a;  +  y)-3/ 


82  SCHOOL    ALGEBRA. 

108.  Stun  and  Difference  of  Two  Cubes.  By  performing 
the  division,  we  find  that 

^±^=o?-ah-\-h\  and   ^^  =  a' J^  ah -\- h\ 
a-{-o  a—  0 

Hence, 

KuLE  2.  The  sum  of  the  cubes  of  two  numbers  is  divisible 
by  the  sum  of  the  numbers,  and  the  quotient  is  the  sum  of 
the  squares  of  the  numbers  minus  their  product. 

Rule  3.  The  difference  of  the  cubes  of  two  numbers  is 
divisible  by  the  difference  of  the  numbers,  and  the  quotient 
is  the  sum  of  the  squares  of  the  numbers  plus  their  proSUct. 

Exercise  25.         (X.  A  '<r|   <>>    ^ 
Write  by  inspection  the  quotient  of  ^      ^  ^  ^ 

l-2x' 


3. 


8. 


l  +  2a; 

21  a' -b' 
Sa  —  b  ' 

27a'  +  b' 
3a-}-b 

64:x'  +  27y' 
4a;+3y 

64a:^-27.y^ 
4a;—  3y 

l-272» 
l-3z  * 

l  +  27z» 
l  +  3z*  ""     2x-4:y' 


9. 

a'b'-c' 
ab~c 

10. 

a'b'  +  c^ 
ab  -{-  c 

11. 

64  +  .y^ 

4  +  y 

12 

343 -8  a' 

7-2a 

1  <l 

8a^  +  ^>' 

2a+b' 

1 1 

x^+729y' 

x'  +  9y 

1  fi 

a' -21b' 

a'~db 

1A 

Sa^-64:y' 

SPECIAL   EULES   OF   DIVISION.  83 

109.   Sum  and  Difference  of  any  Two  Like  Powers.     By 
performing  the  division,  we  find  that 

a' -I' 


a  —  b 
a  +  6 


=  a*  +  a^h  +  a^y  +  ah^  +  6*; 


We  find  by  trial  that 

G?  +  5S  a*  +  5*,  a«  +  h\  and  so  on, 
are  7U)t  divisible  by  a  +  ^  or  by  a  —  &.     Hence, 
If  n  is  any  positive  integer, 

(1)  a" +5**  is  divisible  by  a-\-h  if  n  is  odd,  and  by  neither 
a-\-b  nor  a  —  b  if  n  is  even. 

(2)  a"^  —  h^  is  divisible  by  a  —  b  if  n  is  odd,  and  by  both 
a-{-b  and  a  —  b  if  n  is  even. 

Note.  It  is  important  to  notice  in  the  above  examples  that  the 
terms  of  the  quotient  are  all  positive  when  the  divisor  is  a  —  h,  and 
alternately  positive  and  negative  when  the  divisor  is  a  +  6 ;  also,  that 
the  quotient  is  homogeneous,  the  exponent  of  a  decreasing  and  of  h 
increasing  by  1  for  each  successive  term. 

Exercise  26. 
Find  the  quotient  of 

1.  ^-y','  4.    ^lizl.  7.    ^  +  32 

X  —  y  x-\- 1  a;  4-2 

2.  tizf,  6.    a:* -16  ^     l-m* 


x' 

-1 

X- 

fl 

X* 

-16 

X 

-2 

x^ 

-32 

x-\-y  x  —  2  1  —  m 

x'-l  ^    x^-S2  ^     1  +  m' 

x—1  '     x  —  2  *    1+m 


CHAPTER  VII. 
FACTORS. 

110.  Eational  Expressions.    An  expression  is  rational  when 

none  of  its  terms  contain  square  or  other  roots. 

111.  Pactors  of  Eational  and  Integral  Expressions.  By  fac- 
tors of  a  given  integral  number  m  Arithmetic  we  mean 
integral  numbers  that  will  divide  the  given  number  with- 
out remainder.  Likewise  by  factors  of  a  rational  and  inte- 
gral expression  in  Algebra  we  mean  rational  and  integral 
expressions  that  will  divide  the  given  expression  without 
remainder. 

112.  Factors  of  Monomials.  The  factors  of  a  monomial 
may  be  found  by  inspection.  Thus,  the  factors  of  14a^6 
are  7,  2,  a,  a,  and  h. 

113.  Factors  of  Polynomials.  The  form  of  a  polynomial 
that  can  be  resolved  into  factors  often  suggests  the  process 
of  finding  the  factors. 

Case  I. 

114.  When  all  the  terms  have  a  common  factor. 

(l)  Resolve  into  factors  2x^-\-^xy. 
Since  2 a:  is  a  factor  of  each  term,  we  have 

'2,x^-\'Qxy      2x^  ,  ^xy  ,  q 

%x  2x       2x 

.-.  2x''  +  6xy  =  2x(x-\-Sy). 
Hence,  the  required  factors  are  2x  and  x-{-Sy. 


FACTORS.  85 

(2)  Resolve  into  factors  16  a'  +  4  a^  —  8  a. 

Since  4  a  is  a  factor  of  each  term,  we  have 

16a^  +  4a''-8a^l6a^     4a^     8a 
4a  4a       4a      4a 

=  4a'  +  a  — 2. 

.-.  16a'  +  4a2-8a  =  4a(4a'^  +  a-2). 

Hence  the  required  factors  are  4  a  and  4a^+a— 2. 

Exercise  27. 
Resolve  into  two  factors  : 

1.  Sx^-6a^.  7.  Sa'b-iab\ 

2.  2a''-4a.  8.  8xy  +  4:xy. 

3.  5a5-5a'5l  9.  Sx'-9x'-6x\ 

4.  Sa'-a^  +  a.  10.  8aV- 4a25  +  12ay. 

5.  x'  +  x'i/~xy\  11.  8a'Z)V-4a'^V. 

6.  a*~a'b  +  a'b\  12.  15a':r  —  10aV  + 5aV 

Case  IL 

115,  "When  the  terms  can  he  grouped  so  as  to  show  a  common 
factor. 

(1)  Resolve  into  factors  ac  -}-  ad  -\-  ho  -\^  bd. 

aG-{-ad-\-bc-\-bd=(ac-\-^ad)-\-  {be  +  bd)  (1) 

=  a{c-\-d)-^h{c--\-d)   .  (2) 

=  (a  +  5)(t7  +  c;)>  (3) 

Note.   Since  one  factor  is  seen  in  (2)  to  be  c  +  d,  dividing  by  c  +  d 
we  obtain  the  other  factor,  a  +  6. 


86  SCHOOL   ALGEBRA. 

(2)  Find  the  factors  of  ac  -]-  ad  ~  he  —  hd. 

ac-]-ad—bo~bd=  (ac  +  ad)  —  {be  +  bd) 

=  a{c-\-d)-b{o+d)       , 

=  {a-b){c-\-d). 

Note.  —  Here  the  last  two  terms,  —he  —  bd,  being  put  within  a 
parenthesis  preceded  by  the  sign  — ,  have  their  signs  changed  to  +. 

(3)  Resolve  into  factors  Sar^ -~5a^ —6x +10. 
Ba^-&x'-6x+10  =  (Sx'-5x')-(e>x-10) 

=  x\Sx-b)-2(Sx-5) 
=  (x^-2)(Sx-5). 

(4)  Resolve  into  factors  5  a;^  —  15  ax^  —  x-}-Sa. 
6x^—16aa^—x-{-Sa=(5x^  —  Wax^)  —  (x  —  3a) 

=  6x''(x~Sa)~l(x~Sa) 
=  (5a;2-l)(a;-3a). 

(5)  Resolve  into  factors  6y  —  27a;^y— 10^7  +  45 ar*. 
Q1/-27 x'lZ-lOx+^bx^  =  (4:5a^  -  27 x'y)  -  (lOx  -  6y) 

=  9x'(5x-Sy)-2(5x-S7/) 
=  (9x'-2)(5x-Sz/). 

Note.  By  grouping  the  terms  thus,  {6y  ~27x'^y)  —  {l0x~i5x^), 
we  obtain  for  the  factors,  (2  —  dx'^){3y  —  5x). 

But  {2-9x'^)i3y-5x)  =  {9x^-2){5x-Sy),  since,  by  the  Law 
of  Signs,  the  signs  of  two  factors,  or  of  any  even  number  of  factors, 
may  he  changed  without  altering  the  value  of  the  product. 

Exercise  28. 
Resolve  into  factors  : 

1.  ax--bx-{-ay  —  by.  5.    x'^ -}- ax  ~  bx  —  ah. 

2.  ax  —  bx  —  ay-\-by.  6.    x^  -\- xy  —  ax  — ay. 

3.  ax  —  cy  —  ay-\-cx.  7.   oi?  —  xy  —  ^x-\-^y. 

4.  2ab-Zac  —  2by-\-2>cy.       8.    23?—Zxy+\ax-^ay. 


FACTOES.  87 

9.   fj^h  —  ahx  —  ac-\-  ex.      ^ 

10.  c^hoi}  +  ^'<?^  —  (^cy  —  hc^y, 

11.  3a;^-5y2_6a;'+10a:y^ 

"      12.  ^ax--Vdhx~Vla-\-Voh. 

13.  2a;'  — 3a;' -4a; +  6. 

14.  6a;*  +  8a;'-9a;2_i2a;. 

15.  ax^-\-'bci?  —  ax—h. 

16.  3ca;*-2c?a;'  — 9ca;'  +  6c?a;. 

17.  l  +  15a;*-5a;-3a;'. 

18.  ax^  -\-ci^x-\-a-{-x. 

19.  {a-^h){o-\-d)-Zc{a-\-})). 

20.  (a;-y)'  +  2y(a;-y). 

116.  If  an  expression  can  be  resolved  into  two  equal 
factors,  the  expression  is  called  a  perfect  square,  and  one  of 
its  equal  factors  is  called  its  square  root. 

Thus,  16a;y  =  4a;'y  X  4a;'y.  Hence,  16a;y  is  a  perfect 
square,  and  4a;'y  is  its  square  root. 

/  Note.  The  square  root  of  IBa^y^  may  be  —^^y  as  well  as  +4ic'y, 
for  —  4a;^3/  X  —  ^^y  =  16  a;^^/^  ;  but  throughout  this  chapter  the  posi- 
tive square  root  only  will  be  considered. 

117.  The  rule  for  extracting  the  square  root  of  a  perfect 
square,  when  the  square  is  a  monomial,  is  as  follows : 

{  Extract  the  square  root  of  the  coefficient,  and  divide  the 
index  of  each  letter  hy  2. 

118.  In  like  manner,  the  rule  for  extracting  the  cube 
root  of  a  perfect  cube,  when  the  cube  is  a  monomial,  is, 

Y  Extract  the  cube  root  of  the  coefficient,  and  divide  the  index 
of  each  letter  hy  3. 


88  SCHOOL    ALGEBRA. 

119.  By  §§  99,  100,  a  trinomial  is  a  perfect  square,  if  its 
first  and  last  terms  are  perfect  squares  and  positive,  and  its 
middle  term  is  twice  the  product  of  tlieir  square  roots. 
Thus,  16  o^  —  24  aS  +  95^^  is  a  perfect  square. 

The  rule  for  extracting  the  square   root   of  a  perfect 
square,  when  the  square  is  a  trinomial,  is  as  follows : 
j(f     Extract  the  square  roots  of  the  first  and  last  terms,  and 
connect  these  square  roots  hy  the  sign  of  the  middle  term. 
Thus,  if  we  wish  to  find  the  square  root  of 
16a^-24a5^-95^ 

we  take  the  square  roots  of  16  a^  and  9^^  which  are  4  a 

and  35,  respectively,  and  connect  these  square  roots  by  the 

sign  of  the  middle  term,  which  is  — .     The  square  root  is 

therefore  .         o  z. 

4a  —  60. 

In  like  manner,  the  square  root  of 

16a'  +  24a5  +  95^^  is  4a  +  36. 

Case  III. 

120.  When  a  trinomial  is  a  perfect  square. 

(1)  Resolve  into  factors  x^  -\-2xy-\-  yl 
From  §  119,  the  factors  oi  x"^  -\-2xy -{-  y^  are 

(x-\-y){x-\-y). 

(2)  Resolve  into  factors  x^  —  2x^y  +  y^ 
From  §  119,  the  factors  of  x^  —  2x'^y  -\-  y'  are 

{x''  —  y)(x'-y). 

Exercise  29. 
Resolve  into  factors : 

1.  a'-6a5  +  95^  3.    a'-4a5  +  45^ 

2.  4a2  +  4aZ.  +  6l  4.    a:'^  +  6^y  +  9y^ 


5'ACTOKS.  80 

5.  4:x''-12axi-9a\  13.  i9 x' -  28 xy  +  4.y\ 

6.  a' -10 ah -\-2bb\  14.  1-20^* +  100^1 

7.  4a'-4a  +  l.  15.  81a^  +  l26ab  +  4:9b\ 

8.  49y'-14yz  +  zl  16.  mV  -  16 wwa' +  64 a\ 

9.  a;'— 16a; +  64.  17.  4:0" —  20 ax +  25x'. 

10.  9a:'  +  24a:y-f  16yl  18.    121a' +  198 ay +  81  yl 

11.  16a^-\-8ax-\-x\  19.    a'6 V  -  2 a5 V.r«  +  a:'^ 

12.  25  +  80a;  +  64a;^  20.    49  -  140 >^' +  100 ^J;*. 

Case  IV. 
121.  Wlien  a  binomial  is  the  difference  of  two  squares. 

(1)  Resolve  into  factors  x"^  —  y^. 

From  §  101,    {x ■\- y){x  —  y)  =  x^  —  y'. 

Hence,  the  difference  of  two  squares  is  the  product  of 
twa  factors,  which  may  be  found  as  follows  : 

^  TaJce  the  square  root  of  the  first  term-  and  the  square  root 
of  the  second  term. 

The  sum  of  these  roots  will  form  the  first  factor ; 

The  difference  of  these  roots  will  form  the  second  factor. 

Exercise  30. 
Resolve  into  factors : 


1. 

a^-4. 

6. 

25-16a^ 

11. 

81a;»-4y^ 

2. 

l-x". 

7. 

16  -  25y^ 

12. 

64a*-i\ 

3. 

x'^^f. 

8. 

a'h'  -  1. 

13. 

m'n^  -  36. 

4. 

4a^-49Z)'. 

9. 

:r2-^100. 

14. 

^^-144. 

6, 

^-  4yl 

10. 

121  a'- 36  ^>^ 

15. 

a?  -  25. 

90  SCHOOL    ALGEBRA. 


16. 

49-100y^ 

23. 

49<*-y^\ 

17. 

l-49.2;«. 

24. 

^^a'-U\ 

18. 

4-12l2/«. 

25. 

^la'h'-c\ 

19. 

1  -  169  a^ 

26. 

4:a''c  —  9c\ 

20. 

a'h-'-4:c\ 

27. 

20  aW-bab. 

21. 

^x^-a\ 

28. 

3a^-12aV. 

22. 

^x''-f\ 

29. 

9a'-Slb\ 

30.  25-64yl 

31.  16a;" -9a;/. 

32.  25x''-16a'x\ 

33.  36aV-49a*. 

34.  x'-167/\ 

35.  l-400a;*. 

36.  4:0^0  — 9 c\ 


122.    If  the  squares  are  compound  expressions,  the  same 
method  may  be  employed. 

(1)  Resolve  into  factors  (a;+  3y)^  —  16  a^ 

The  square  root  of  the  first  term  is  x  +  Sy. 

The  square  root  of  the  second  term  is  4  a. 

The  sum  of  these  roots  is  ic  +  Sy  +  4  a. 

The  difference  of  these  roots  is  a;  +  By  —  4  a. 

Therefore  {x  +  Syf  -  IQa^  =  {x  +  3y  +  4:a)  {x  +  3y  - 4:a). 

(2)  Resolve  into  factors  a^  —  (3  5  —  5  c)^ 

The  square  roots  of  the  terms  are  a  and  (36  —  5  c). 
The  sum  of  these  roots  is  a  +  (3  6  —  5  c),  or  a  +  3  6  —  5  c. 
The  difference  of  these  roots  is  a  —  (3  6  —  5  c),  or  a  —  3  Z>  +  5  c. 
Therefore  a^  -  (3  6  -  5  c)^  =  (a  +  3  &  -  5c)(a-  3  6  +  5c). 

/ 123.  If  the  factors  contain  like  terms,  these  terms  should 
be  collected  so  as  to  present  the  results  in  the  simplest 
form. 

(3)  Resolve  into  factors  (3a  +  bbf  —  (2a  —  2>,b)\ 

The  square  roots  of  the  terms  are  3  a  +  5  6  and  2  a  —  3  6. 
The  sum  of  these  roots  is  (3  a  +  5  6)  +  (2  a  —  3  6), 

or3a  +  56  +  2a  — 36  =  5a  +  2&. 
The  difference  of  these  roots  is  (3  a  +  5  6)  —  (2  a  —  3  6), 

or3a  +  56  —  2a  +  36  =  a  +  86. 
Therefore  (3 a  +  5  6)2  -  (2a  -  3  6)2  =  (5 a  +  2 &) (a  +  8  6). 


FACTORS.  91 

Exercise  31. 
Resolve  into  factors : 

1.  (x  +  yy-z\  11.  (a-by-(c-d)\       >^ 

2.  (:r-3/)^-2^  12.  (2a-\-by-25c\  ^ 

3.  {x~2yj-^z\  13.  {x-\-2yf  —  {^x-yf. 

4.  (a  +  3^>)^-16e^  14.  {x-^Zy-i^x-^J.      ^^ 

5.  a;=^-(y-zy.  15.  {a^h- cj -(a-h  -  cf. 

6.  a^- (35 -2^7.  16.  (a-3a;y-(3a-2a;y. 

7.  y'-{2a-\-Z6)\  17.  (2a-l)^-(3a+l)'. 

8.  \-{x-\-^l)\  18.  (a;-5)^-(a;  +  y-5y. 

9.  9a'^-(a;-3c)^  19.  (2a+5-c)^-(a-26  +  cy. 
10.  16a^-(22/-3z)».  20.  (a  +  25-3c7-(a  +  5c)l 

124.  By  properly  grouping  the  terms,  compound  expres- 
sions may  often  be  written  as  the  difference  of  two  squares, 
and  the  factors  readily  found. 

(1)  Resolve  into  factors  c^  —  2ah-\-h'^  —  ^  c^. 

=  (a- by -9c' 

=  (a-b  +  Sc)(a-b-3c). 

(2)  Resolve  into  factors  12ab  +  9x' —  4:a' -9 b\ 

Note.  Here  12  o6  shows  that  it  is  the  middle  term  of  the  expres- 
sion which  has  in  its  first  and  last  terms  a^  and  5^,  and  the  minus 
sign  before  A^a^  and  96^^  shows  that  these  terms  must  be  put  in  a 
parenthesis  with  the  minus  sign  before  it,  in  order  that  they  may  be 
made  positive. 

The  arrangement  will  be 

9x'-{^a'-l2ab  +  9b'')  =  9x'-(2a-Uy 

=  (3a;+2a-3i)(3a;— 2a+36). 


92  SCHOOL  ALGEBRA. 

(3)  Resolve  into  factors  —  a^  +  6^  —  c^  +  c?^  +  2  ac  +  2  5c?. 
Note.   Here  2ac,  2bd,  and  —a',  —  c',  indicate  the  arrangement 
required. 

=  (b'  +  2bd+  d')  -(a'-2ac  +  (^) 

=  {bJ^dy-{a-cf 

=  (6  +  c?+a-c)(6  +  <i-a  +  c). 

Exercise  32. 
Resolve  into  factors : 
l.(a^-\-2ab  +  h^-(Ac\  5.    a^ - x^ -y"" —  2xy.     -^ 

2.  'V  -2xy-^f  '-9a\  i-     ^.    \-a^ -2ab -b\    l^ 

3.  5^-a;^  +  4aa7-4a^^       7.    a'  +  Z>^  +  2a5  -  16a26^ 

4.  4a'  +  4a5  +  Z>'-a:l  8.    4a'- 9a' +  6a- 1.^^^ 

9.  a'  +  5'  — c'  — c?'-2a5  — 2cd 

10.  x^-\-y''-2xy-2ab-a^  —  b^. 

11.  9a;'-6a;+l— a'-4a5-4^>'. 

12.  a'  +  2a^>-a;'-6a;y-9y'  +  5'. 

13.  a;'-2a;  +  l-Z>'  +  25y-y'. 

14.  9-6a;  +  a;'-a'-8a5-166'. 

15.  4-4a;4-a;'-4a-l-4a'. 

.    16.    a*-a'-9  +  5*  +  6a-2a'^»'. 

125.  A  trinomial  in  the  form  of  a*  +  a'5'  +  Z>*  can  be 
written  as  the  diiference  of  two  squares. 

Since  a  trinomial  is  a  perfect  square  when  the  middle 
term  is  twice  the  product  of  the  square  roots  of  the  first 
and  last  terms,  it  is  obvious  that  we  must  add  a'6'  to  the 
middle  term  of  a*  +  c^b^  +  6*  to  make  it  a  perfect  square. 


FACTORS.  93 

We  must  also   subtract   d?h^  to  keep  the  value   of  the 
expression  unchanged.     We  shall  then  have 

(1)  a*  +  a^6'-{-^>*  =  a*+2a^52^Z.*-a'5» 

=  (a^  -\-ah-\-  h')  (a^  -ab  +  b'). 

If  in  the  above  expression  we  put  1  for  5,  we  shall  have 

(2)  a'  +  a'  -{-1  =  {a'  +  20"  +  1)  -  a" 

=  (a'  +  1  +  a)  (a'  +  1  -  a) 
=  (a^  +  a+l)(a'-a+l). 

(3)  Resolve  into  factors  4^*  — 37a;y  +  9y*. 

Twice  the  product  of  the  square  roots  of  4:X^  and  dy*  is  12x'^y'^. 
We  may  separate  the  term  —^^x^y"^  into  two  terms,  —12x^y'^  and 
—  25x^y^,  and  write  the  expression 

(4a;*  -  12a:y  +  9y*)  -  25a;y 

=  (2x''-^7/y-2bxy 

^(2x'+bx7/-Sy'){2x'-5x2/-Sy'). 

Exercise  33. 
Resolve  into  factors : 

1.  x*  +  xy  +  y\  6.    9a*+26a'5^  +  255*. 

2.  x'-i-x'^  +  l.  7.    4:x*-21xy  +  9y'. 

3.  9a*-15a'  +  l.  8.    4a*- 29aV  +  25c*. 

4.  16a*-17a'  +  l.  9.    4a*+ 16aV  +  25c*. 

5.  4a*"-13aHl»  10.    25a;*+ 31a;y  +  16y*. 


94  SCHOOL    ALGEBRA. 

U^^    -^78".  Case  V. 

126.  When  a  trinomial  lias  the  form  x^  +  ax  +  b. 

From  §  104  it  is  seen  that  a  trinomial  is  often  the  product 
of  two  binomials.  Conversely,  a  trinomial  may,  in  certain 
cases,  be  resolved  into  two  binomial  factors. 

127.  If  a  trinomial  of  the  form  x^  -\-ax-{-h  is  such  an 
expression  that  it  can  be  resolved  into  two  binomial  fac- 
tors, it  is  obvious  that  the  first  term  of  each  factor  will  be 
X,  and  that  the  second  terms  of  the  factors  will  be  two 
numbers  whose  product  is  h,  the  last  term  of  the  trinomial, 
and  whose  algebraic  sum  is  a,  the  coefficient  of  x  in  the 
middle  term  of  the  trinomial. 

(1)  Resolve  into  factors  o:^  +  lla;  +  30. 

We  are  required  to  find  two  numbers  whose  product  is  30  and 
whose  sum  is  11. 

Two  numbers  whose  product  is  30  are  1  and  30,  2  and  15,  3  and 
10,  6  and  6 ;  and  the  sum  of  the  last  two  numbers  is  11.     Hence, 

x''  -\-llx  -\-2>0  =  {x  -\-b){x  -\-^). 

(2)  Resolve  into  factors  x"^  —  1  x-{-12. 

We  are  required  to  find  two  numbers  whose  product  is  12  and 
whose  algebraic  sum  is  —  7. 

Since  the  product  is  +  12,  the  two  numbers  are  hoth positive  or  hoth 
negative ;  and  since  their  sum  is  —  7,  they  must  both  be  negative. 

Two  negative  numbers  whose  product  is  12  are  —  12  and  —1,-6 
and  —2,-4  and  —3 ;  and  the  sum  of  the  last  two  numbers  is  —7. 
Hence, 

x''-1x-\-  l2  =  {x-4:){x-2>). 

(3)  Resolve  into  factors  x"^  -\-2x  —  2^. 

We  are  required  to  find  two  numbers  whose  product  is  —  24  and 
whose  algebraic  sum  is  2. 


FACTOES.  95 

Since  the  product  is  —  24,  one  of  the  numbers  is  positive  and  the 
other  negative ;  and  since  their  sum  is  +  2,  the  larger  number  is 
positive. 

Two  numbers  whose  product  is  —  24,  and  the  larger  number  posi- 
tive, are  24  and  —  1,  12  and  —  2,  8  and  —  3,  6  and  —  4  ;  and  the  sum 
of  the  last  two  numbers  is  +  2.     Hence, 

a;*  +  2a:  -  24  =  (a;  +  6)  (a;  -  4). 

(4)  Resolve  into  factors  a;'  —  3  a;  —  18. 

We  are  required  to  find  two  numbers  whose  product  is  —  18  and 
whose  algebraic  sum  is  —  3. 

Since  the  product  is  —18,  one  of  the  numbers  is  positive  and  the 
other  negative ;  and  since  their  sum  is  —  3,  the  larger  number  is 
negative. 

Two  numbers  whose  product  is  — 18,  and  the  larger  number  nega- 
tive, are  —  18  and  1,-9  and  2,-6  and  3  ;  and  the  sum  of  the  last 
two  numbers  is  —  3.     Hence, 

a^  -  3ar-  \^  =  {x~  6)(a:  +  3). 

(5)  Resolve  into  factors  x^  —  lOary  -j-  9y'. 

We  are  required  to  find  two  expressions  whose  product  is  9^^  and 
whose  algebraic  sum  is  —  lOy. 

Since  the  product  is  +93/^,  and  the  sum  —  lOy,  the  last  two  terms 
must  both  be  negative. 

Two  negative  expressions  whose  product  is  9  y'',  are  —  9y  and  —y, 
—  Sy  and  —  83/;  and  the  sum  of  the  first  two  expressions  is  —  lOy. 
Hence, 

oc^—  10a;y  +  93/^=  (x  —  9y){x  —  y). 


Exercise  34. 

Resolve  into  factors : 

1.  x'-^Qx+lb.  4.   x^-Sx-10. 

2.  x^-Sx+15.  5.    x'-\-5ax  +  6a\ 

3.  a;''  +  2a:— 15.  6.    x^—5ax-\-6a\ 


96  SCHOOL   ALGEBRA. 

7.  x'-^x-lb,  32.  x''  +  ax-6a\      /- 

8.  x^-\-6x  +  6.  >3».  x'-ax  —  6a\    /- 

9.  x^-6x+Q.  34.  x'  +  6x7/  +  4:f.  ^ 

10.  x^  +  x-6.  35.  x''--^x2/-4.y\  ^ 

11.  a;' -a? -6.  36.  x""  —  b x^  +  4:i/\   ^ 

12.  a;' +  6^7 +  5.  37.  x^^Zxy  —  ^y". 

13.  ^''  —  Gaj  +  S.  38.  a;'  +  3a;y  +  22/'. 

14.  a:'  +  4a;--5.  39.  a'^  -  7 a5  +  10 ^>l /- 

15.  x^-^x—b.  40.  aV  — 3aa:-54.  ,1^ 

16.  ^''  +  9^  +  18.  41.  x'-7x-4A.       i^ 

17.  x'-9x  +  18.  42.  a^'^  +  aj-lSa     ^£_ 

18.  x'  +  Sx-lS.  43.  :r'-15a;  +  50. 

19.  ii7*--3a?— 18.  44.  a'^  — 23 a +120. 

20.  a7»  +  9a:+8.  45.  a' +  17a -390. 

21.  x^~Qx-\-8.  46.  c'^ +  25^-150. 

22.  x''  +  1x-S.  47.  c'  — 58^  +  57. 

23.  x^-1x-8.  48.  a*-lla^^'^  +  30^»^ 

24.  a;2+7:c+10.  49.  z'  +  9zy  +  20y\ 

25.  a;»-7a7+10.  50.  xy-}-19xyz  +  i8z^. 

26.  a;'  +  3a;— 10.  51.  a*'Z>'- 13a5c  +  22c^ 

27.  x'-5ax  —  b0a\  52.  a'-16a5-366l 

28.  x'y^-Sxy  —  A.  53.  a;'^+ 17:ry  +  30^^ 

29.  a'-3aa;-54ri;2^  54.  x''~7xy~18y\ 

30.  a«--a»<?  — 2c^  55.  c''+c-20. 

31.  y'-28yz+187z'.  56.  a'' +  16a2>- 260S'. 


FACTORS.  . 

57. 

y'-5y2-84zl 

58. 

a;^- 11  a; -152. 

59. 

lS-Sx-x'  =  ~(x'  +  Sx-lS). 

60. 

SS-}-8x-x'  =  ~(a^-Sx-  33). 

61. 

IS-lx-af. 

97 


Case  VI. 
128.   When  a  trinomial  has  the  form  ax'^  +  bx  +  o^ 
From  §  105, 

(3a; -.2)  (5  a: +  3) 

=  15a:^+9a;-10a:-6  =  15a;''-a;-6.      (1) 
(8a:-2)(5a;-3) 

=  15a;^-9a:-10a;+6  =  15a:'-19a:+6.  (2) 

Consider  the  resulting  trinomials  : 

Thie  first  term  in  (1)  and  (2)  is  the  product  3  a;  x  5  a;. 

The  middle  term  in  (1)  is  the  algebraic  sum  of  the  products 

3xxS  and  {-2)x5x. 

The  middle  term  in  (2)  is  the  algebraic  sum  of  the  products 

3a;x(-3)and(-2)x5a;. 

The  last  term  in  (1)  is  the  product  (—  2)  X  3. 
The  last  term  in  (2)  is  the  product  (-  2)  x  (-  3). 
The  trinomials  have  no  monomial  factor,  since  no  one  of  their 
factors  has  a  monomial  factor.     Hence, 

1.  If  the  third  term  of  a  given  trinomial  is  negative, 
the  second  terms  of  its  binomial  factors  have  Unlike  signs. 

2.  If  the  third  term  is  positive,  the  second  terms  of  its 
binomial  factors  have  the  same  sign,  and  this  sign  is  the 
sign  of  the  middle  term. 


98  SCHOOL   ALGEBRA. 

3.  If  a  trinomial  has  no  monomial  factor,  neither  of  its 
binomial  factors  can  have  a  monomial  factor. 

(1)  Resolve  into  factors  6a;'+ 17a;+ 12. 

The  first  terms  of  the  binomial  factors  must  be  either  6  x  and  x,  or 
3a;  and  2x. 

The  second  terms  of  the  binomial  factors  must  be  12  and  1,  or  6 
and  2,  or  3  and  4. 

We  therefore  write 

I.   (6a;  +     ){x+    ),     or    II.   (3a;  +     )(2a;+     ). 

For  the  second  terms  of  these  factors  we  must  reject  1  and  12 ;  for 
12  put  in  the  second  factor  of  I.  would  make  the  product  6  a;  x  12  too 
large,  and  put  in  the  first  factor  of  I.,  or  in  either  factor  of  XL,  the 
result  would  show  a  monomial  factor. 

We  must  also  reject  6  and  2;  for  if  put  in  I.  or  II.  the  results 
would  show  monomial  factors ;  and  for  the  same  reason  we  must 
reject  3  and  4  for  I. 

The  required  factors,  therefore,  are  (3  a;  +  4)  and  (2  a;  +  3). 

(2)  Resolve  into  factors  14ar^  —  11a;  —  15. 

For  a  first  trial  we  write 

(7a;     ){2x    ). 

Since  the  third  term  of  the  given  trinomial  is  — 15,  the  second 
terms  of  the  binomial  factors  will  have  unlike  signs,  and  the  two 
products  which  together  form  the  middle  term  will  be  one  +.  and 
the  other  — .  Also,  since  the  middle  term  is  —11  a;,  the  negative 
product  will  exceed  in  absolute  value  the  positive  product  by  11  x. 

The  required  factors,  therefore,  are  (7  a;  +  5)  and  (2  a;  —  3). 

Exercise  35. 

Resolve  into  factors : 

1.  2a;'4-5a;+3.  4.    24a;' -  2a:y- 15^. 

2.  3a;' -a; -2.  5.    36a;'- 19a;y- 6y'. 

3.  5a;' -8a;  4- 3.  6.    15a;'+ 19.ry  +  6/. 


FACTORS.    (  7  y     '  )  99 

7.  6a:' +  7a; +  2.  14.    15s^ -26xy i-Sy\ 

8.  6x'-x  —  2.  15.    9x^-i-6xy  —  Sy\ 

9.  Ibx^+Ux-S.  16.    6x'-xy-S5y\ 

10.  8a;' -10a; +  3.  17.  10a;'- 21a;y  -  lOy'. 

11.  18a;' +  9a; -2.  18.  14a;'- 55 a;y  + 21 3/'. 

12.  24a;'  — 14a:y  — 5y'.  19.  6a;' —  23a;y +  20^'. 

13.  24a;'-38a;y+15y'.  20.  6a:'  + 35a;y- 6y'. 

Case  VII. 
129.   When  a  binomial  is  the  sum  or  difference  of  two  cubes. 

From  §108,     ^-±4^  =  a^ -a5  +  ^»^ 

and  ^^JZ^^a'  +  ah  +  b'; 

a  —  0 

.-.  a'  +  b'  =  (a  +  b)(a'-ab  +  b') 

and  .         a'-b'=(a-b)(a'  +  ab  +  b'). 

In  like  manner  we  can  resolve  into  factors  any  expres- 
sion which  can  be  written  as  the  sum  or  the  difference  of 
two  cubes. 


(1)  Resolve  into  factors  8a'  +  275^    (-^-^SX^'^l^ 

Since  by  §  118,  8  a^  -  (2  a)' ,  and  276«  =  (3i')',  we  can 
write  8  a'  +  27  b'  as  (2  af  +  (3  by. 

Since         »         a''  +  b^  =  (a  +  b)  (a'  -  ab -{-  b^), 

we  have,  by  putting  2a  for  a  and  35'  for  b, 

(2a)»  +  (35')'  =  (2a  +  36')(4a'-6a5'+950- 


:  fcfr^^-  -^ 


100  SCHOOL    ALGEBRA. 

(2)  Eesolve  into  factors  125^'  — 1.    (s'/^Tf'fjU^X  -^  ^  JS^' 

125a;^-l  =  (5a;)'-l 

(3)  Resolve  into  factors  x^  -\-  y^. 

=^{x^^f){x'~xy  +  y% 

130.    The  same  method  is  applicable  when  the  cubes  are 
compound  expressions. 

(4)  Resolve  into  factors  {x  —  3/)^.+  z^. 
Since       a^  +  5'  =  (a  +  h)  {o?  -ah-{-  b'), 

we  have,  by  putting  :?;  — y  for  a  and  z  for  b, 

{x-yy+z'  =  [(x-y)+z][(x-yy-(x-y)zi-z'] 

=  (x-y  +  z)(x''  —  2xy-}-y^-xz+yzi-z''). 

Exercise  36. 
Resolve  into  factors  : 


1. 

a'-\-8b\ 

5. 

2:10^ 

y-1.              9.    2l6a'-P. 

2. 

a'-27a\ 

6. 

a'  +  27b\              10.   64:a'- 27 b\ 

3. 

a'  +  64. 

7. 

xy 

-64.               11.    343 -a;l 

4. 

125a«+l. 

8. 

64:  a' 

'+125b\         12.   a'J'  +  343. 

13.    8a'- 

-b\ 

19.    Sx'-{x-yf. 

14.    216^ 

m^  + 

n\ 

20.    8(x-{-yy  +  z\ 

15.    (a  +  bf- 

-1. 

21.    729y'-64:z\ 

16.    (a- 

hy+i. 

22.    (a  +  by-(a-by. 

17.  (2x-\-yy-(ix-yy,    23.    729a«  +  216c^ 

18.  I -(a -by.  24.    a7y-512z^. 


FACTORS.  101 

131.  We  will  conclude  this  chapter  by  calling  the  stu- 
dent's attention  to  the  following  statements  : 

1.  When  a  binomial  has  the  form  rp**  —  y**,  but  cannot  be 
written  as  the  difference  of  two  perfect  squares,  or  of  two 
perfect  cubes,  jt  is  still  possible  to  resolve  it  into  two 
rational  factors,  one  of  which  is  x — y.      Thus  (§  109), 

2.  When  a  binomial  has  the  form  :r"-|-?/",  but  cannot  be 
written  as  the  sum  of  two  perfect  cubes,  it  is  still  possible 
to  resolve  it  into  two  rational  factors,  except  when  n  is 
2,  4,  8,  16,  or  some  other  power  of  2.     Thus  (§  109), 

a^J^h'={a-\-h){a'-a%^a^b''-ab^-\-h''). 

But  a^-^-h^,  a^-\ri>^,  a^-\-b^,  cannot  be  resolved  into 
rational  factors. 

3.  The  student  must  be  careful  to  select  the  best  method 
of  resolving  an  expression  into  factors.  Thus,  a^  —  b^  can 
be  written  as  the  difference  of  two  squares,  or  as  the  dif- 
ference of  two  cubes,  or  be  divided  by  a  —  Z>,  or  by  a  +  h. 
Of  all  these  methods,  the  best  is  to  write  the  expression  as 
the  difference  of  two  squares,  as  follows 

{aj  -  {by  =  {a'  +  b')  (a«  -  b') 

=  (a+bXa'-ab-j-b'){a-b)(a'+ab+b'). 

4.  From  the  last  example,  it  will  be  seen  that  an  expres- 
sion can  sometimes  be  resolved  into  three  or  more  factors. 

a^-h'  =  (x'-\-b*)(x'-h') 

=  (x'  +  b*)(x'  +  b')(x'-b') 

=-(x'-}-b*){x'  +  b')(x-i-b)(x-b). 


102  SCHOOL   ALGEBRA. 

5.  When  a  factor  occurs  in  every  term  of  an  expression, 
this  factor  should  first  be  removed.     Thus, 

Sx'-bOa''-{-  4:x-10a  =  2(4.x^  -  2ba'  +  2x -  6a) 

=  2[{4.a^  -  26a')  +  (2x  -  5a)] 
=  2(2a7-5a)(2:f  +  5a+l). 

6.  Sometimes  an  expression  can  be  easily  resolved  if  we 
replace  the  last  term  but  one  by  two  terms,  one  of  which 
shall  have  for  a  coefficient  an  exact  divisor  or  a  multiple  of 
the  last  term.     Thus, 

(1)  x'-6x'  +  llx-16  =  (s^-6x'  +  6x)  +  {6x-16). 

=x{x^-5x+6)-\-6(x-S) 
=  (x-S)[x(x-2)  +  6] 
=  (x-S)(x^-2x-}-6). 

(2)  a;'-9a;^+26a:-24  =  (:zr^-9.^^  +  14a;)  +  (12a;-24) 

=  x(x'-9xi-l4:)  +  12(x-2) 
=x(x-7)(x-2y-\-l2(x-2) 
=  (x-2)(x'-1x+12) 
=  (x-2)(x-S)(x-4:). 

(3)  a^-26x-6  =  (x'  -25x)--(x-}-  5) 

=  x(x'-2b)-(x-i-6) 
=  (x-{-6)(x''-5x-l). 

(4)  3^-{-Bx'-4:  =  (x'  +  2x')-{-(x'-4:) 

=  x'(x  +  2)  +  {x'-4:) 
=  (x-{-2)(x'-i-x-2) 
=  (x  +  2){x  +  2)(x-l). 


FACTORS.  103 


Exercise 

37. 

Examples  for 

Review. 

Resolve  into  factors : 

1. 

a'  — 9a. 

23. 

12a:^-a:-l. 

2. 

xY-^xy'-Zxy. 

24. 

12a:'^-a:-20. 

3. 

;r'  +  a;'  +  a;+l.  — 

25. 

9a^+12a  +  4. 

4. 

x'-^y-\-2x-xy. 

26. 

a?-h''-c'-{-2bc. 

5. 

3a;'  +  2a;^-9:r-6. 

27. 

a:*  +  a:y  +  y\ 

6. 

7. 

^r^- 14a: +  49. 
36a;^-492/^ 

28. 
29. 

2'-6z-40.                      ^ 

8. 

x^-y\ 

30. 

a'- 19a +  84. 

9. 

{x-yy-h\     , 

31. 

a:'^  +  2aa:  +  3^'a:  +  6a6. 

10. 

^'  +  2/^. 

32. 

a:^  +  m'  — n'^  — 2ma:. 

11. 

x'-f. 

33. 

4a:*-a:^ 

12. 

x'-\-y\ 

f  34. 

x'^  +  y'\ 

13. 

x'-{a-h)\ 

35. 

9a:*  +  21a:2^H25y*. 

14. 

rn?-\-2mn^n^-l. 

36. 

x'-^  +  y^^2xy. 

15. 

o?-{m-\-  n)\ 

37. 

2a:»+3a:y-22/^ 

16. 

a:^- 11a; +  18. 

38. 

2a'-7a  +  6. 

17. 

^2_|_4^_45 

39. 

a:*-7a:'+l. 

18. 

a;' +  13a: +  36. 

40. 

l-a'-h'-2ah. 

19. 

ar^-13a:-48. 

41. 

3a:*-6a:^  +  9a:2 

20. 

a:'  +  9a:  — 36. 

42. 

^3  _  5^2  _  2^  .^10. 

21. 

10  a:' +  a: -21. 

43. 

x^-]-ax~bx  —  ah. 

22.    6a:'-a:-12.^  44.    2a:'- 3a:y  +  4aa:  -  6ay. 


104  SCHOOL   ALGEBRA. 

45.  ax^-\-hx^  —  ax—h.  63.  6a'  — a— 77. 

46.  a^  +  5'  +  a  +  5.  64.  5^?*  -  15c^  -  90cl 

47.  d^ —  h^ -\- a  —  h.  65.  a^x  — c^x -\- o}y  —  c^y. 

48.  {x-yj  —  2y{x-y).  66.  16a;*  — 81. 

49.  l-10a;y  +  25a;y.  67.  a;*  +  ^'  +  l. 

50.  a'^-&2  +  2Z>e-cl  68.  27a;^-64a^ 

51.  a;'^  +  4?/'^  — z'^- 4^;?/.  69.  x^-\-y^. 

52.  a'-4^>'-9c'  +  12^c.     70.  :r^-y^ 

53.  4^'  +  9y'-z'-12a;y.  71.  a«-256. 

54.  {a-^hy-{c-d)\  72.  ^*+ 16aV  +  256a*. 

55.  ^r^  +  y''.  73.  \-{x  —  yJ. 

56.  32^^  — cl                        ?74.  {x^yf^{^x-yj. 
67.  a^  +  64yl  75.  :r^  -  216. 

58.  729-07*.  76.  Zx^-\-x  —  % 

59.  x'''-y^\  77.  2~-3a-2al 

60.  (a  +  5)*-l.  78.  4  — 5c-6cl 

61.  a'^-J'  +  a-S.  79.  2xy  -  x"  -  y"" -\- z\ 

62.  a'  +  a  +  35-9^)l  80.  4a*  —  9a'^  + 6a  -  1. 

81.  a'  — 2a5  +  ^>'  +  12a;y-4a;'~9y^ 

82.  2a;''  — 4:i;yH-22/'  +  2aa;  — 2ay. 

83.  {a  +  hy-l-ah{a-\-h  +  l). 

84.  0:^-0;'  + 3a; +  5.     (See  §131,  6.) 

85.  60:^-230;' +  1607 -3.     (See  §131,  6.) 

86.  a;'  +  y''  +  z''-2a;2/-2a;z  +  2yz.     (See  §  103.) 
/87.  ^.a'lf-ia'-l-h'-cy. 


CHAPTER  VIII. 

COMMON  FACTORS  AND  MULTIPLES. 

132.  Oommon  Factors.  A  common  factor  of  two  or  more 
integral  numbers  is  an  integral  number  which  divides  each 
of  them  without  a  remainder. 

133.  A  common  factor  of  two  or  more  integral  and  rational 
expressions  is  an  integral  and  rational  expression  which 
divides  each  of  them  without  a  remainder. 

Thus,  5  a  is  a  common  factor  of  20  a  and  25  a  ;  3  x^y^  is 
a  common  factor  of  12a;yand  Ibx^y^. 

134.  Two  numbers  are  said  to  be  prime  to  each  other 
when  they  have  no  common  factor  except  1. 

135.  Two  expressions  are  said  to  be  prime  to  each,  other 
when  they  have  no  common  factor  except  1. 

136.  The  highest  common  factor  of  two  or  more  integral 
numbers  is  the  greatest  number  that  will  divide  each  of 
them  without  a  remainder. 

137.  The  highest  common  factor  of  two  or  more  integral 
and  rational  expressions  is  an  integral  and  rational  expres- 
sion of  highest  degree  that  will  divide  each  of  them  with- 
out a  remainder. 

Thus,  2)0^  is  the  highest  common  factor  of  2>a^,  ^a^, 
and  12  a*  ;  bx^y^  is  the  highest  common  factor  of  lOx^y^ 
and  15ar^2/^. 

For  brevity,  we  use  H.  C.  F.  to  stand  for  "highest 
common  factor." 


106  SCHOOL    ALGEBRA. 

To   find  the  highest   common   factor  of  two   algebraic 
expressions : 

Case  I. 

138.   When  the  factors  can  be  found  by  inspection. 

(1)  Find  the  H.  C.  F.  of  ^^a^h"  and  md}h\ 

^2a'b'  =  2.x  ^Xlxaaaxbb; 
eOa'b*  =  2x2x3x5xaaX  bbbb. 
.-.  the  H.  C.  F.  -=  2  X  3  X  aa  X  bb,  or  6a'b\ 

(2)  Find  the  H.  C.  F.  of2a'xi-2ax^  and  ^abxy-\-^bx'y. 

2a^x  -{-2ax^        =  2  ax  (a -{- x) -, 
3  abxT/  +  3  bx'^y  =  3  bxy  (a  +  x). 
.-.  theH.O.F.         =x(a-\-x). 

(3)  FindtheH.O.F.  of4a;^  +  4a;-48,  6a;'- 48a;  +  90. 

4a;2+    4a;-48  =  4(a;'  +  a;-12) 

=  4(a;-3)(a;  +  4); 

6a;»  -  48a;+  90  =  6 (a;'  -  8a;  +  15) 
=  6(a;-3)(a;-5); 
.-.theH.O.F.         =2(a;-3) 
=  2a;-6. 

Hence,  to  find  the  H.  0.  F.  of  two  expressions : 

Mesolve  each  expression  into  its  sijnplest  factors. 
^  Find  the  product  of  all  the  common  factms,  taking  each 
comm.(m  factor  the  least  number  of  times  it  occurs  in  any  of 
the  given  expressions.  • 


COMMON    FACTORS   AND    MULTIPLES.  107 

Exercise  38. 
FindtheH.C.F.  of 

1.  120  and  168.  4.    36aV  and  28^*^. 

2.  36a:' and  27 :i-*.  5.    48a'^>'c  and  60aV. 

3.  42aVand60aV.  6.    8(a -f- ^>)^  and  6(a  + 5)'. 

7.  l2a{x-\-yy  2.ndiU{x-\-y)\ 

8.  {x-iy{x-\-2Y  2.ndi{{x-2>)(x-\-2)\] 

9.  24a^6'(a  +  ^>)and42a-^5(a  +  5)l 

10.  x\x-2>y  ^ndix'-^x.      12.    a;'^  —  4a;  and  ^'- 6a;  +  8. 

11.  .^'^-le  anda:'  +  4:r.         13.    a;'^  — 7:r+12  and  a;'-16. 

14.  ^x^-^y""  2,ndi\2x'  —  xy  —  ^y\ 

15.  a:'— 7  a;  — 8  and  0:'  + 5a;  4- 4. 

16.  x''-\-?>xy-lOy''  ^ndix'-2xy-2>by\ 

17.  a;*  — 2a;'-24a;2and6a;^-6a;*-180a;'. 

18.  a;'  — 3a;V  anda;'-27yl 
19..  1  + 64a;' and  l-4a;  +  16a;^ 
20.  a;*-81  anda;*  +  8a;'  — 9.     • 

21.  a;'  +  2a;-8,  a:»4:jra?4-12. 

22.  a;'  — 6a:  +  5,  a;'4-3.r-40. 

23.  3a*  +  lba?h  -  na'h\  6a» -  30a»5  +  36a5'. 

24.  6a;V-12a;y^+6y',  3a;y  +  9a;y»-12y*. 

25.  1--16C*,  l  +  c'-12c*. 

26.  8a;'  +  2a;-l,  6ar»  +  7a;  +  2. 

27.  6a;'  +  ar-2,  12a;^  — a;-6. 

28.  15a;»-19a;V  +  6a;y«,  lOx' -  a^y-Za^f. 

29.  10a:»y+9a;'y'  — 9a;y',  ^xy*  +16y»-4a;*y. 


108  SCHOOL   ALGEBRA. 

Case  II. 

139.  When  tlie  factors  cannot  be  found  by  inspection. 

The  method  to  be  employed  in  this  case  is  similar  to 
that  of  the  corresponding  case  in  Arithmetic.  And  as  in 
Arithmetic,  pairs  of  continually  decreasing  numbers  are 
obtained,  which  contain  as  a  factor  the  H.  0.  F.  required, 
so  in  Algebra,  pairs  of  expressions  of  continually  decreas- 
ing degrees  are  obtained,  which  contain  as  a  factor  the 
H.  0.  F.  required. 

140.  The  method  depends  upon  the  following  principles: 

(1)  Any  factor  of  an  expression  is  a  factor  also  of  any 
multiple  of  that  expression. 

Thus,  if  c  is  contained  3  times  in  A,  then  c  is  contained 
9  times  in  3  A,  and  3  m  times  in  m  A. 

(2)  Any  common  factor  of  two  expressions  is  a  factor  of 
their  sum,  their  difference,  and  of  the  sum  or  dfference 
of  any  m^ultiples  of  the  expressions. 

Thus,  if  c  is  contained  5  times  in  A,  and  3  times  in  B, 
then  c  is  contained  8  times  in  A-\-  B,  and  2  times  in 
A-B. 

Also,  in  5^  +  2^  it  is  contained  5x5  +  2x3,  or  31  times, 
and  in  5  ^  —  2  ^  it  is  contained  5x5  —  2x3,  or  19  times. 

(3)  The  H.C.F.  of  two  expressions  is  not  changed  if  one 
of  the  expressions  is  divided  by  a  factor  that  is  not  a  factor 
of  the  other  expression,  or  if  one  is  onuUiplied  hy  a  factor 
that  is  not  a  factor  of  the  other  expression. 

Thus,  the  H.  C.  F.  of  4  a^bc"^  and  aVc?  is  not  changed  if 
we  remove  the  factors  4  and  b  from  ^a^bc^,  and  d  from 
a^c^d\  or  if  we  multiply  ^a^bc^  by  7,  and  aVc?  by  11. 


COMMON    FACTORS    AND    MULTIPLES.  109 

141.  We  will  first  find  the  greatest  common  factor  of 
two  arithmetical  numbers,  and  then  show  that  the  same 
method  is  used  in  finding  the  H.O.F.  of  two  algebraic 
expressions. 

Find  the  greatest  common  factor  of  18  and  48. 

18)48(2 
36 

12)18(1 
12 

6)12(2 
12 


Since  6  is  a  factor  of  itself  and  of  12,  it  is,  by  (2),  a  fac- 
tor of  6+12,  or  18. 

Since  6  is  a  factor  of  18,  it  is,  by  (1),  a  factor  ol  2  X  18, 
or  36  ;  and  therefore,  by  (2),  it  is  a  factor  of  36  +  12,  or  48. 

Hence,  6  is  a  common  factor  of  18  and  48. 

Again,  every  common  factor  of  18  and  48  is,  by  (1),  a 
factor  of  2  X  18,  or  36  ;  and,  by  (2),  a  factor  of  48  -  36, 
or  12. 

Every  such  factor,  being  now  a  common  factor  of  18  and 
12,  is,  by  (2),  a  factor  of  18-12,  or  6. 

Therefore,  the  greatest  common  factor  of  18  and  48  is 
contained  in  6,  and  cannot  be  greater  than  6.  Hence  6, 
which  has  been  shown  to  be  a  common  factor  of  18  and  48, 
is  the  greatest  common  factor  of  18  and  48. 

142.  It  will  be  seen  that  every  remainder  in  the  course  of 
the  operation  contains  the  greatest  common  factor  sought ; 
and  that  this-is  the  greatest  factor  common  to  that  remain- 
der and  the  preceding  divisor.     Hence, 

The  greatest  common  factor  of  any  divisor  and  the  corre- 
sponding dividend  is  the  greatest  common  factor  sought. 


110  SCHOOL    ALGEBRA. 

143.  Let  A  and  J5  stand  for  two  algebraic  expressions, 
arranged  according  to  the  descending  powers  of  a  common 
letter,  the  degree  of  B  being  not  higher  than  that  of  A. 

Let  A  be  divided  by  JB,  and  let  Q  stand  for  the  quo- 
tient, and  H  for  the  remainder.     Then 

-B)A(Q 
BQ 

B 

Whence,  E  =  A- BQ,  and  A  =  BQi-B. 

Any  common  factor  of  B  and  JR  will,  by  (2),  be  a  factor 
of  BQ  +  B,  that  is,  of  A  ;  and  any  common  factor  of  A  and 
B  will,  by  (2),  be  a  factor  of  ^  —  BQ,  that  is,  of  B. 

Any  common  factor,  therefore,  of  A  and  B  is  likewise 
a  common  factor  of  B  and  B.  That  is,  the  common  fac- 
tors of  A  and  B  are  the  same  as  the  common  factors  of  B 
and  B ;  and  therefore  the  H.  0.  F.  of  B  and  B  is  the 
H.C.F.  of  ^and^. 

If,  now,  we  take  the  next  step  in  the  process,  and  divide 
B  by  B,  and  denote  the  remainder  by  jS,  then  the  H.  0.  F. 
of  S  and  B  can  in  a  similar  way  be  shown  to  be  the 
same  as  the  H.O.  F.  of  B  and  B,  and  therefore  the  H.  C.  F. 
of  A  and  B  ;  and  so  on  for  each  successive  step.     Hence, 

The  H.C.F.  of  any  divisor  and  the  corresponding  divi- 
dend is  the  IT.  C.  F.  sought. 

If  at  any  step  there  is  no  remainder,  the  divisor  is  a  fac- 
tor of  the  corresponding  dividend,  and  is  therefore  the 
H.C.F.  of  itself  and  the  corresponding  dividend.  Hence, 
the  last  divisor  is  the  H.  C.  F.  sought. 

•  Note.  From  the  nature  of  division,  the  successive-remainders  are 
expressions  of  lower  and  lower  degrees.  Hence,  unless  at  some  step 
the  division  leaves  no  remainder,  we  shall  at  last  have  a  remainder 
that  does  not  contain  the  common  letter.  In  this  case  the  given 
expressions  have  no  common  factor. 


COMMON    FACTORS   AND    MULTIPLES.  Ill 

Find  the  H.C.F.  of  2a;2  +  a;-3  and  A:i^+&x'  —  x-6. 

6a;' +  5a; -6 
6a;'  +  3a;-9 

2a;-f|-3)2a;'+    a;-3(a;— 1 
2a;'  +  3a; 

-2a;-3 
.-.  the  H.C.F.  =  2a;  +  3.  -2a; -3 

Each  division  is  continued  until  the  first  term  of  the  remainder  is 
of  lower  degree  than  that  of  the  divisor. 

144.  This  method  is  of  use  only  to  determine  the  com- 
pound factor  of  the  H.  C.F.  Simple  factors  of  the  given 
expressions  must  first  be  separated  from  them,  and  the 
H.  0.  F.  of  these  must  be  reserved  to  be  multiplied  into  the 
compound  factor  obtained. 

Find  the  H. C.F.  of 

12a;*  +  30a;^-72a;'  and  32a;=^  + 843;^^- 176a;. 

l^x'  +  30a;'  -  72a;2  =  6a;'(2a;'  +  5a;  -  12). 
32a;'  +  84a;^  -176a:=  4a;(8a;^  +  21a;  -  44). 
6  a;''  and  4  a;  have  2  a;  common. 

2  a;'  +  5  a;  -  1 2)  8  a;'  +  21  a;  -  44  (4 
8a;' +  20  a; -48 


x+,    4)2a;'  +  5a;-12(2a;-3 
2a;'  +  8a; 

-3a;-12 
.-.  theH.C.F.  =  2a;(a;-|-4).  -3a;-12 


112  SCHOOL   ALGEBRA. 

145.    Modifications  of  this  method  are  sometimes  needed. 

(1)  Find  the  H.C.F.  of  4:x''-8x-5  and  12x'-4:X-65. 

4a;'-8^-5)12:r^-    4a;- 65(3 
12x'~24:x-15 


20  X-  50 

The  first  division  ends  here,  for  20a;  is  of  lower  degree  than  4:X^. 
But  if  20  a;  — 50  is  made  the  divisor,  ix^  will  not  contain  20  a;  an 
integral  number  of  times. 

The  H.  C.  F.  sought  is  contained  in  the  remainder  20  a;  —  50,  and  is 
a  compound  factor.  Hence  if  the  simple  factor  10  is  removed,  the 
H.C.F.  must  still  be  contained  in  2a;  — 5,  and  therefore  the  process 
may  be  continued  with  2  a;  —  5  for  a  divisor. 

2a;-5)4:r^-    8a:-5(2:r+l 
4:x'  —  10x 


2x-b 
2x-5 

.-.  theH.G.F.  =  2a:-5. 

(2)  Find  the  H.C.F.  of 

21:i;^  -  4a;'  -  15a;  -  2  and  21  x'  -32x''~  54a;  - 

2lx^  -  4a;'  -  15a;  -  2)  21  x'  -  32a;'  -  54a;  -  7(1 
21a;^-    4a;' -15a; -2 


-28a;'-39a;-5. j  ^  ^^ 

The  difficulty  here  cannot  be  obviated  by  removing  a  simple  factor 
from  the  remainder,  for  —  28 a;'-^-  39a;  —  5  has  no  simple  factor.  In 
this  case,  the  expression  21  a;^  —  4 a;^—  15a;  —  2  must  be  multiplied  by 
the  simple  factor  4  to  make  its  first  term  exactly  divisible  by  —  28  3;^^. 

The  introduction  of  such  a  factor  can  in  no  way  affect  the  H.  C.  F. 
sought,  for  4  is  not  a  factor  of  the  remainder. 

The  signs  of  all  the  terms  of  the  remainder  may  be  changed ;  for 
if  an  expression  A  is  divisible  by  —F,  it  is  divisible  by  +  F. 

The  process  then  is  continued  by  changing  the  signs  of  the  re- 
mainder and  multiplying  the  divisor  by  4. 


COMMON    FACTORS   AND   MULTIPLES.  113 

2Sx'  +  S9x  +  6)S4:x'~    16a;^-    60a;-    8(3a; 
84:x^  +  U7x'+    15x 

-433a;^—   75a;-    8 
Multiply  by —4,         —4 

532a;^  + 300  a; +  32(19 
532a;^+741a;  +  95 

Divide  by  -  63,  -  63)- 441a; -63 

7a;+    1 

7a;+ l)28a;'  +  39a;  +  5(4a;  +  5 
28a;' +    4a; 

35a7  +  5 
.-.  the  H.  C.  F.  =  7a;  +  1.  35a; +  5 

(,3)  Find  the  H.C.F.  of 

8a;2  +  2a;  -  3  and  6a;'  +  5a;2  —  2. 


6a;3  4-    5a;'--    2 
4 

E;'4-2a;-3)24a;^  +  20a;'-    8   (3 
243;^+    6a;'-    9a; 

a;+7 

14a;' +    9a;- 
Multiply  by  4,           4 

•   8 

56a;' +  36a; - 
56  a;' +  14a;- 

32 

21 

Divideby  11,                 ll)22a;- 

2a;- 

•11 

•    l)8a; 
8a; 

'+2a;- 
•'-4a; 

-3(4a;+3 

.-.  tlieH.C.F.  =  2a;-l. 

6a;- 
6a;- 

-3 
-3 

114 


SCHOOL    ALGEBRA. 


The  following  arrangement  of  the  work  will,  be  found 
most  convenient : 


e>x-s 

6a?-3 


6a^  + 
4 


5x'~    2 


24a;^  +  20a;^-   8 
24a;' +    6x'~    9x 


8 


14^7^4-    9a;- 

J 

66x^-{-SQx~S2 

56a;''  +  14a;-21 

ll)22a;-ll 

2x-    1 


Sx 


+  7 


4a;  +  3 


146.  From  the  foregoing  examples  it  will  be  seen  that,  in 
the  algebraic  process  of  finding  the  H.  C.F.,  the  follow- 
ing steps,  in  the  order  here  given,  must  be  carefully- 
observed  : 

I.  Simple  factors  of  the  given  expressions  are  to  be  re- 
moved from  them,  and  the  H.  G.  F.  of  these  is  to  be  reserved 
as  a  factor  of  the  H.  0.  F.  sought. 

II.  The  resulting  compound  expressions  are  to  be  ar- 
ranged according  to  the  descending  powers  of  a  common 
letter ;  and  that  expression  which  is  of  the  lower  degree  is 
to  be  taken  for  the  divisor;  or,  if  both  are  of  the  same 
degree,  that  whose  first  term  has  the  smaller  coefiicient. 

III.  Each  division  is  to  be  continued  until  the  remainder 
is  of  lower  degree  than  the  divisor. 

IV.  If  the  final  reflaainder  of  any  division  is  found  to 
contain  a  factor  that  is  not  a  common  factor  of  the  given 
expressions,  this  factor  is  to  he  removed;  and  the  resulting 
expression  is  to  be  used  as  the  next  divisor. 

V.  A  dividend  whose  first  term  is  not  exactly  divisible 
by  the  first  term  of  the  divisor,  is  to  be  multiplied  by  such 
a  number  as  will  make  it  thus  divisible. 


COMMON    FACTORS    AND    MULTIPLES.  115 

Exercise  39. 

Find  by  division  the  H.  0.  F.  of 

1.  ^x'  +  Sx-lO,  4:a^  +  7x''-Sx-W. 

2.  2x'-6x'  +  5x--2,  8x''~2Sx'-{-17x~6, 

3.  20x'  +  2x'  —  lSx  +  4:8,  20^' -  17a;'^  +  48a;- 3. 

4.  4:x'-2x'~16x-91,  12a^-2Sx''~S7x-^2. 

5.  12a;' +  4:^2  + 17a: -3,  24^;^  -  52a;^  +  14a;- 1. '^ 

6.  2a;'  +  5a;^~2a;4-3.  3a:' +  2a;^- 17a;+ 12. 

7.  Sx^~i)a^~'X'  +  l6x-2b,  4a7^4- 7a:'- 3  a;  -  15. 

8.  4a;' -4a:' -5a: +  3,  lOa^'- 19a:  +  6. 

9.  6a;'- 133:^  +  33:^'+ 2a;,  6a;*  -  lOar' +  4a;' -  6a;-f  4. 

10.  2a;*  — 3a:'  +  2a:'-2a:-3,  4a:*  + 3a;='  + 4a:-3. 

11.  3a:*-a;'-2a:''  +  2a;-8,  6a;' +  13a;^  + 3a;+20. 

12.  3a:'  +  2a:*  +  a;^  3a:*  + 2a:' —  3a;' +  2a:  -  1. 

13.  6ar^-9a:*+lla:'+6a:'-10a:,  4a;^+10a;*+10a;'4-4a:'+60a:. 

14.  2a;5-lla:'-9,  4a;^+lla:*  +  81. 

15.  a:*  — 4a;'  +  10a:'— 12a;+9,  a;*  +  2a;'  +  9. 

16.  2a:'-3a:'-16a:  +  24,  4a:'  +  2a;*- 28a:'- 16a:'-32a:. 

17.  a;*-a:'-14a:'  +  a:  +  l,  a:^  -  4a;*  -  ar' -  2a;' +  8a:  + 2. 

18.  6a:'  — 14aa:'4-6a'a:  — 4  a',  a:*  —  aa:' —  a V  —  a'a;  —  2  a*. 

19.  2a*-2a'  — 3a'  — 2a,  3a*-a'-2a'- 16a. 

20.  2a:'+7aa;'  +  4a'a;— 3a',  4 a;' +  9 aa;' -  2 a'a;  -  a". 

21.  2a;'-9aar'  +  9a'a;-7a',  4a:'~  20  aa;'  + 20  a'a;- 16  a'. 

22.  2a:*-{-9a;'  +  14a;  +  3,  3a:*  +  14a;'  + 9a;4- 2. 

23.  9a;'^— 7a:^4-8a;'+2a;  — 4,  6a:*  -  7a:'— 10a:'  +  5a:+2. 


116  SCHOOL    ALGEBRA. 

147.  The  H.  0.  F.  of  three  expressions  may  be  obtained 
by  resolving  them  into  their  prime  factors ;  or  by  finding 
the  H.  C.  F.  of  two  of  them,  and  then  of  that  and  the  third 
expression. 

For,  if  A,  B,  and  (7  are  three  expressions, 

and  D  the  highest  common  factor  of  A  and  B, 
and  U  the  highest  common  factor  of  I)  and  O, 

Then  D  contains  every  factor  common  to  A  and  B, 
and  U  contains  every  factor  common  to  B  and  O. 

.'.  E  contains  every  factor  common  to  A,  B,  and  C, 

Exercise  40. 
Find  the  H.  0.  F.  of 

1.  a:'  +  3a;  +  2,  x^-{-^x  +  2>,  a;'  +  6:r  +  5. 

2.  x'-^x-lO,  x'-1x-2>0,  x^-llx+lO. 

3.  a^-l,  x^-2x''-\-\\  0^-2x^1. 

4.  &x^  +  x-2,  2x^  +  7x-4:,  2x^-1x  +  S. 

5.  a''  +  2ab  +  h\  a'-b',  a' +  20^6  i- 2 ab' +  b\ 

6.  x'-5ax-{-4:a\  x'-3ax  +  2a\  Sx'-10ax+1a\ 

7.  x'-}-x  —  6,  x'-2x'-x  +  2,  x'  +  Sx'  —  6x~8. 

8.  x^+7x'-^5x-l,  x''+Sx-3x^—l,  Sx^+^x'+x  —  l. 

9.  x'-6x'-^llx-e>,  r^;^-8^^+19a:-12,  x^-9x'+26x-24:. 

148.  Common  Multiples.  A  common  multiple  of  two  or 
more  numbers  is  a  number  which  is  exactly  divisible  by 
each  of.  the  numbers. 

A  common  multiple  of  two  or  more  expressions  is  an 
expression  which  is  exactly  divisible  by  each  of  the  ex- 
pressions. Thus,  48  is  a  common  multiple  of  4,  6,  and  8 ; 
4:S(x'^  —  '}/)  is  a  common  multiple  of  3(a;  — y)  and  8(a7  +  y). 


COMMON   FACTORS   AND    MULTIPLES.  117 

149.  The  lowest  common  multiple  of  two  or  more  numbers 
is  the  least  number  that  is  exactly  divisible  by  each  of  the 
given  numbers. 

The  lowest  common  multiple  of  two  or  more  expressions 
is  the  expression  of  lowest  degree  that  is  exactly  divisible 
by  each  of  the  given  expressions.  Thus,  2^{x^—y'^)  is  the 
lowest  common  multiple  oi2){x—y)  and  ^{x-{-y). 

We  use  L.C.M.  to  stand  for  "lowest  common  multiple." 
To  find  the  L.'C.  M,  of  two  or  more  algebraic  expressions : 

Case  I. 

150.  When  the  factors  of  the  expressions  can  be  found  by 
inspection. 

(1)  Find  the  L.C.M.  of  ^^a'h"  and  e>Oa^b\ 

42a'52---2x3x  IXa^Xh'; 
ma?h'  =2  k2  X  3  X  5  X  a2  X  b\ 
The  L.C.M.  must  evidently  contain  each  factor  the  greatest  num- 
ber of  times  that  it  occurs  in  either  expression. 

.-.  L.  C.  M.  =  2  X  2  X  3  X  7  X  5  X  a''  X  ^*, 

^^20a?b\ 

(2)  Find  the  L.C.M.  of 

A:x'  +  4a:-  48,  6^;'^-  48:i;  +  90,  4.x^  -  10a:  -  6. 

4a:^+   4^-48=4(0:^ +.0:- 12)    =  2x  2(a:- 3)(a:  +  4)  ; 
6^:^-48  :i;  +  90  =  6(a;^- 8a:  rf  15)  =  2x3(^-3)  (a:- 5); 
^x^-lOx-   6  =  2(2a^^-5a:-3)-2(:r-3)(2a:+l). 
.•.L.C.M.  =  2x2x3x(a;-3)(a:+4)(a:-5)(2a:+l). 

Hence,  to  find  the  L.  C.  M.  of  two  or  more  expressions : 

Hesolve  each  expression  into  its  simplest  factors. 

Find  the  product  of  all  the  different  factors,  taking  each 

factor  the  greatest  number  of  times  it  occurs  in  any  of  the 

given  expressions. 


118  SCHOOL    ALGEBRA. 

Exercise  41. 
Find  the  L. CM.  of 

1.  24,  32,  and  60. 

2.  24 aV,  60 aV,  and  32  aV. 
.3.  x^  —  'Ixy  -^-y^  and  x^  —  y^. 

4.  ^2__4^_^4^  ^2_j_4^_^4^  and  a;'  — 4. 

5.  :^^  +  a'  and  :r^  —  al 

6.  x^ -\- ax -\- a^ ^  x'^  —  o?,  and  :r' —  a^ 

7.  x\x  —  Zy  2.ndi  x^~bx-\-Q. 

8.  :i;'+7rr+12  and  x'~'dx\ 

9.  :^'2_Y^_|_2o,  ^2_4^_5^  ^^2_^_2. 

10.  l-3.r-4a;^  l-4:r-5:^;^  l-9a;-102;'. 

11.  6a?'  +  7;ry-3/,  2>x''-\- II xy-4:y\  2x'^llxy-{-l2y\ 
.     12.  8  — 14a  +  6a^  4a  +  4a'-3a^  4a'  +  2a^  — 6a*. 

13.  6^^  +  72;2-3:r,  3.r^  +  14:r-5,  6 3;^  + 39:?; +  45. 

14.  6a2:  +  9Z':r — 2ay  —  3^?/,  6:r^  +  3aa; — 2xy  —  ay. 

15.  12aa;  — 9ay-8:ry  +  6y^  6a:r  +  3ay  — 4.^?/  — 22/^ 

16.  21x^-a\  e>x^+ax-a\  \bx''-bax^^bx~ah. 

17.  :i-'-l,  2:r'-^-l,  3:r'-^-2. 


Case  II. 

151.  When  the  factors  of  the  expressions  cannot  be  found  by 
inspection. 

In  this  case  the  factors  of  the  given  expressions  may  be 
found  by  finding  their  H.  C.  F.  and  dividing  each  expres- 
sion by  this  H.  C.  F. 


COMMON    FACTORS   AND    MULTIPLES. 


119 


Find  the  L. CM.  of 

6a;'-lla;V  +  2y'and  9.27'-22a:y^-8y». 


Qa^-Ux'^  +2y^ 


9  07^-22  ^y^-  8i/' 
2 


ISx^-Uxy'-lQy' 


lly)SSx'y--Uxf-22y' 


Sx^  —  ^xy  ~  ^y"^  2x—y 
:.  theH.C.F.  =  3;r^-4a;y-2yl 
Hence,  ^x''~llx''y-\-2y^  =  {2x—y)(2>x''-^xy-2y^), 
and  ^x^-22xy''-8y'  =  (^x-\-^y){?>x''-^xy-2y''). 

:.  the  'L.GM.  =  {2x-y){^x-\-^y){^x'-4:xy-2y''), 


Exercise  42. 
Find  the  L. CM.  of 

1.  ^x^-1ax'-20a'x,  3a;'  +  aa7  — 4al 

2.  3:r'-13a:'  +  23.r-21,  6a:'  +  a;' -  44a;  + 21. 

3.  3;r^  —  3:r^y  +  xy^  —  y^  4a;'  —  :?;^y  —  2>xy'^. 
■  4.  ^*_2c'  +  c,  2c'  —  2c^-2c-2.     ' 

5.  :r'-8a:  +  3,  r?;^-3a;^-f  21^7-8 

6.  a^  —  ^o?x-^l2ax''-8x\  20^ -8ax -\-8x\ 

7.  2a7'  +  a;'-12:f  +  9,  2r''  — 7a;'  +  12a7-9. 

8.  lx^-~2x''-b,  lx^+l2x''+l0x-\-b. 


9.   a;*— 13a;' +  36,  x' 


1x''-\-x+Q. 


10.  2a;'  +  3a;'-7:r— 10,  4a;-''  — 4a;'  — 9a;  +  5. 

11.  12a;'-a;'-30a;-16,  6a;' -  2a;' -  13a;- 6. 

12.  6a;'  +  a.'2-5a;-2,  6a;' +  5a;' ~  3a;- 2. 

13.  a;'-9a;'  +  26a;~24,  a;'-12a;'  +  47a;-fia^ 

ilVERSITY   i 


ilVERSITY    i 


CHAPTER  IX. 

FRACTIONS.    . 

152.  An  algebraic  fraction  is  the  indicated  quotient  of 
two  expressions,  written  in  the  form  ^• 

0 

The  dividend  a  is  called  the  numerator,  and  the  divisor  b 
is  called  the  denominator. 

The  numerator  and  denominator  are  called  the  terms  of 
the  fraction. 

153.  The  introduction  of  the  same  factor  into  the  divi- 
dend and  divisor  does  not  alter  the  value  of  the  quotient, 
and  the  rejection  of  the  same  factor  from  the  dividend  and 
divisor  does  not  alter  the  value  of  the  quotient.      Thus 

12^3   2x12 _ 3    12_4^ ^ 3      j^  follows,  therefore,  that 
4  2X4  4-f-2 

The  value  of  a  fraction  is  not  altered  if  the  nwtnerator  and 
denominator  are  both  multiplied,  or  both  divided,  by  the 
same  factor. 

Reduction  of  Fractions. 

154.  To  reduce  a  fraction  is  to  change  its  form  without 
altering  its  value. 

Case  I. 

155.  To  reduce  a  fraction  to  its  lowest  terms. 

A  fraction  is  in  its  lowest  terms  when  the  numerator  and 
denominator  have  no  common  factor.  We  have,  therefore, 
the  following  rule  : 


FRACTIONS.  121 

JResolve  the  numerator  and  denominator  into  their  prime 
factors,  and  cancel  all  the  common  factors ;  or,  divide  the 
numerator  and  denominator  by  their  highest  common  factor . 

Keduce  the  following  fractions  to  their  lowest  terms  : 
38a^Z>V     2xl9a^5V     25V 


(1) 
(2) 


bla^bc''      3xl9a^5c^        3  a 

a^  —  o(^  _{a  —  x)  (a^  -j- ax -}- x"^)  _  a"^ -\- ax -\- cc* 
d^  —  x"^  {a  —  x){a-\-x)  a-\-x 


^       a'^  +  Sa  +  e       (a +  3)  (a +  2)      a  +  S' 

,^^    e)x'-bx~^  _(2a;-3)(3:g  +  2)_3a;  +  2 
^  ^  ^x^-2x-\b      (2a; -3)  (4a; +  5)      4a;  +  5" 

.  .  a;^  — 4a;^  +  4a;-l 
^  ^  a?-2x'-\-^x-^ 

"We  find  by  the  method  of  division  the  H.  0.  F.  of  the 
numerator  and  denominator  to  be  ir  —  1. 

The  numerator  divided  by  a;  —  1  gives  x^  —  2>x-\-l. 
The  denominator  divided  by  a;  —  1  gives  x^  —  x-\-Z. 

a;^  —  4^;^^  +  4a;  —  1  _  a;'^  —  3a;  +  1 
'■  a;^-2a;'  +  4a;-3       a;^-a;  +  3' 


Exercise  43. 

Eeduce  to  lowest  terms  : 

^     6a6^  4     42m^5  ^     34aa;y 

^a^h  '    ^Ldmn""'  '    bla'x/ 

Sab'c  SOxy'z*  35aW 


Iba'bV  ISxyz"  baWc 

26a^y^  6     ^^^'^'  9     58a&V 

'    2>^x/  *   28mV*  .  *   87  a W 


122  SCHOOL    ALGEBRA. 


^^      9a:y-12y^  ^^     Ax'' +  12  ax +  9  a' 

12a;*' -16^  *          8a;' +  27  a' 

11      4a^-9c^  2Q     a;^-y^-2.yz-z' 

4a=^  +  6a(?  *    a;^  +  2a;y +.3/^-2'^ 

*    a' +  4a +  4  *         3^  —  y^, 

13        ^'-5&                .  22     2a'^+17a  +  21 

^4        20(a'-e')  23     (^  +  ^y-^^ 

4(a'4-ac  +  c^)            ■  '     (a4-5  +  c/ 

15.  —^±£ 24.  ^'-y'. 

a;'  +  2a:y  +  y'  3/  — a; 

16.  ^-^-^^  .  25.  (-+^y-^\ 

x''  +  2x-lb  (x  +  hj-a^ 

x^-^x  +  \b  26     (^  +  ^)^-(^  +  ^)' 

2a;^-13a;  +  21  *    (a  +  c)^- (6  +  c^y 

a;^-a;-20  .  (a  +  c)^  -  &^ 

2a;^-7a;-15  '    4aV  -  (a^  +  c"- ^7 


Keduce  by  finding  the  H.  C.  F.  of  the  terms  : 

,      a;'  — 6a7-4  ^^     3ar>  +  17a;'  +  22a;  +  8 

6a;'4-25a;^  +  23a;+6' 

a;'  — 3a;'—  15a? +  25 
a;'  +  7a;^  +  5a;-25* 

34     2a;'  +  a;'-8a;  +  3 
3ar'  +  8a;'  +  a;-2 

.^     ar'  +  4a;2-8a;+24 
a;*-a;»  +  8ar-8 


AiO. 

3a;'-8a;  +  8 

29. 

a;'-3a;  +  2 
a;'  +  4:^-''-5 

30. 

3a;»-a;^-a;-l 

3ar»-4a;'^-a;  +  2 

^1 

a;* -13a;' +  36 

iz;*-ar'-7a7'  +  a;+6 

33. 


FBACTIONS.  123 


Case  II. 


156.   To  reduce  a  fraction  to  an  integral  or  mixed  expression. 

x^  —1 

(1)  Reduce' —  to  an  integral  expression. 

X  ~1 

•      t^  =  a^^a:+l.     (§108) 
X  —  1 

x^  —  1 

(2)  Reduce to  a  mixed  expression. 

'  X  -p  J. 


a? 

-1 

b+1 

x' 

+  x' 

x'  -x+1 

-x'- 

-1 

-x'- 

—  X 

x-1 

x  +  l 

-2 

s^- 

-1 

1   1 

=  x'- 

-+i     '■ 

x+l  x+1 

Note.  By  the  Law  of  Signs  for  division, 

^and_^-  =  — 1-. 
X  +  l  — (a;+l)         x  +  l 

The  last  form  is  the  form  usually  written. 

157.  If  the  degree  of  the  numerator  of  a  fraction  equals 
or  exceeds  that  of  the  denominator,  the  fraction  may  be 
changed  to  a  mixed  or  integral  expression  by  the  following 
rule  : 

Divide  the  numerator  by  the  denominator. 

Note.  If  there  is  a  remainder,  this  remainder  must  be  written  as 
the  numerator  of  a  fraction  of  which  the  divisor  is  the  denominator, 
and  this  fraction  with  its  proper  sign  must  be  annexed  to  the  integral 
part  of  the  quotient. 


124  SCHOOL    ALGEBRA. 

Exercise  44. 
Reduce  to  integral  or  mixed  expressions : 

1, •  o. • 

4:x  a  +  b 

„     3a;2-9a;-2  ^     Sx'-\~2x-{-l 


10. 


4.    •"    '  .y.  11. 


12. 


6. ^.  13. 


7.    i — —•  14.    - 


X-\-4: 

a'-^2x' 
a  +  2x' 

4a;^-3.T-54 

X  —  4: 

2^-3a 

•4a^-5 
a'^  +  a  -  2 


Case  III. 
158.   To  reduce  a  mixed  expression  to  a  fraction. 

The  process  is  precisely  the  same  as  in  Arithmetic.  Hence, 
Multiply  the  integral  expression  by  the  denominator,  to 

the  product  add  the  numerator,  and  under  the  result  write 

the  denominator. 

(1)  Reduce  to  a  fraction -\-  x  —  b. 

X  —  4 

^-^  I  X     5_^-3  +  (^-4)(rr-5) 
a;  — 4  :r  — 4 

^a:- 3 +  ^'-9^  +  20 

X  —  4: 

_x''  —  ^x^-ll 

X  —  4: 


FRACTIONS.  125 

(2)  Keduce  to  a  fraction  a-b-  ^^  ~  ^^  ~  ^'. 

a  +  b 

^      ^     a'-ab-b'_(a-b)(a  +  b)-(a'-ab-b^) 
a-{-b  a-j-b 

_a'-b^-a'  +  ab  +  b^ 
a-i-b 

ab 

~a-{-b 

Note.  The  dividing  line  between  the  terms  of  a  fraction  has  the 
force  of  a  vinculum  affecting  the  numerator.  If,  therefore,  a  minus 
dgn  precedes  the  dividing  line,  as  in  Example  (2),  and  this  line  is 
removed,  the  numerator  of  the  given  fraction  must  be  enclosed  in  a 
parenthesis  preceded  by  the  minus  sign,  or  the  sign  of  every  term  of 
the  numerator  must  be  changed. 

Exercise  45. 
Keduce  to  a  fraction  : 

1.  .  +  l  +  i^.  8.   .  +  4-^. 

2x  x—2> 

2.  a; -1-1-^^-^.  9.    a^-ax^x^--  ^ 


3.  a  +  b- 

4.  x-1- 

5.  1 


2x 

2ab 
a  +  b 

2 
x-2 

a  —  b 


a-\-x 

10. 

a'  + 

ax-{- 

a  —  x 

1 1 

a-dx 

a-{-2x. 

4 

12. 

3a- 

-2b- 

Sa'-2b' 

a-\-b  a-{-  b 

6.   5a-l+M.  13.   2:.-7-2kLl3^. 


a 


I          a'  +  x"                     1.     K^      Q  ,  3a;  +  21 
7.    a-{-x ! 14.    5a;  — d-j !— — • 

a  +  x  x-{-  7 


126  SCHOOL   ALGEBRA. 

Case  IV. 
159.   To  reduce  fractions  to  tlieir  lowest  common  denominator. 

Since  the  value  of  a  fraction  is  not  altered  by  multiply- 
ing its  numerator  and  denominator  by  the  same  factor 
(§  153),  any  number  of  fractions  can  be  reduced  to  equiva- 
lent fractions  having  the  same  denominator. 

The  process  is  the  same  as  in  Arithmetic.  Hence  we 
have  the  following  rule  : 

Find  the  lowest  common  m^ultiple  of  the  denominators  ; 
this  will  be  the  required  denominator.  Divide  this  deno7ni- 
nator  by  the  denominator  of  each  fraction. 

Multiply  the  first  numerator  by  the  first  quotient,  the  sec- 
ond numerator  by  the  second  quotient,  and  so  on. 

The  products  will  be  the  respective  numerators  of  the 
equivalent  fractions. 

Note.  Every  fraction  should  be  in  its  lowest  terms  before  the 
common  denominator  is  found. 

(1)  Reduce    - — -,  -^,  and    — —  to  equivalent  fractions 
4:a^    6  a  6  a 

having  the  lowest  common  denominator. 

The  L.  C.  M.  of  4  a\  3  a,  and  6  a^  =  12  a\ 
The  respective  quotients  are  3  a,  4a^  and' 2. 
The  products  are  9  ax,  8  a^y,  and  10. 
Hence,  the  required  fractions  are 

9  ax     8  ah/    ^^^     10 
12  a^'    12  a^'  12  a^ 


(2)  Reduce , , to 

^^  x^+5x-}-6     x'  +  4:X-}-S     x'-{-2x+l 

equivalent   fractions  having  the  lowest  common   denom- 
inator. 


^ 


FRACTIONS.  127 

2  3 


a^  +  5a?  +  6    a;2  +  4a;  +  3     a;^  +  2a;  +  1 

1  2  3 


(a?  +  3)(a;  +  2)     (a:  +  3)(a;  +  l)     (a;  +  l)(a;  +  l) 
.*.  the  lowest  common  denominator  (L.  C.  D.)  is 

(a;  +  3)(a;  +  2)(cp  +  l)(a;  +  l). 
The  respective  quotients  are 

{x  +  l){x  +  1),  (a;  +  2)(x  +  1),  and  {x  +  3)  (a;  +  2). 
The  respective  products  are 

l{x  +  l){x  +  1),  2(a;  +  2)(a:  +  1),  and  3(a;  +  3) (a  {■  2). 

Hence  the  required  fractions  are 

(a?  +  l)(a;  +  l)  2(a;  +  2)(a;  +  1) 

{x  +  S){x  +  2){x  +  If     (x  +  3)ix  +  2)  (x  +  If 

3(.T  +  3)(a;  +  2) 
(a;  +  3)(a;  +  2)(a;+l/ 

Exercise  46. 
Express  with  lowest  common  denominator : 


^     a  —  2x    Sx''  —  2ax 
Sa              9  ax 

^     Aa'  +  c'     2a-\~c 
'    Aa'-c''    2a-c 

„    1 

2 
x  +  S 

5      ^'  +  y'           1 

"•   .  +  2' 

25a;^-4y^'    5:^^  +  2y 

3         " 

a' 

g     a:  +  2     07-2 

a;  — a 

x'  -  a' 

07—2'     07  +  2 

7.       1     , 

X  +  1/ 

1 

2 

^-y 

^'-y" 

8      '1      . 

1 

1-4 

3 

*•   1  +  2^' 

07^'      l-207 

•■rf?' 

7 

3 

3+57 

3-07 

1 

1 

07^-907+18      07' -1007  +  24 


128  SCHOOL   ALGEBRA. 


Addition  and  Subtraction  op  Fractions. 

160.  The  algebraic  sum  of  two  or  more  fractions  which 
have  the  same  denominator,  is  a  fraction  whose  numerator 
is  the  algebraic  sum  of  the  numerators  of  the  given  frac- 
tions, and  whose  denominator  is  the  common  denominator 
of  the  given  fractions.  This  follows  from  the  distributive 
law  of  division. 

If  the  fractions  to  be  added  have  not  the  same  denomi- 
nator, they  must  first  be  reduced  to  equivalent  fractions 
having  the  same  denominator.    (§  159.) 

To  add  fractions,  therefore. 

Reduce  the  fractions  to  equivalent  fractions  having  the 
same  denominator ;  and -write  the  sum  of  the  numerators 
of  these  fractions  over  the  common  denominator. 

161.  When  the  denominators  are  simple  expressions. 
(1)  Simplify-^ 3^  +  -l^- 

The  L. CD.  =  12. 

The  multipliers,  that  is,  the  quotients  obtained  by  dividing  12  by 
4,  3,  and  12,  are  3,  4,  and  1, 

Hence  the  sum  of  the  fractions  equals 

9a-126     8a-46  +  4c  .  a-4c 


12 

12 

'      12 

9a 

-12b- 

-(8a 

-45 

+  4c)  +  a- 

-4c 

12 

9o 

-12b. 

-8a  +  46- 

-4c  +  a- 

-4c 

12 

2a 

-86- 
12 

•8c 

a  — 

46-4 

c 

FEACTIONS.  129 

The  above  work  may  be  arranged  as  follows : 

The  L. CD.  =  12. 

The  multipliers  are  3,  4,  and  1,  respectively. 

3(3a-46)      =     9a  — 126  =  1st  numerator. 

—  4(2a  —  6  +  c)  =  —  8a+    46  —  4c=2d  numerator. 

l(a  —  4c)         =        a  — 4c=3d  numerator. 

2a-    86-8c 
or    2  (a  —  4  6  —  4  c)  =  the  sum  of  the  numerators. 

p.      ,.  2(a-46-4c)     a-46-4c 

.*.  sum  of  fractions  =  -^^ — ^  = 

12  6 


Exercise  47. 
Simplify : 

1     ^~1      x—S     x—7 . x—2 
'       2  5  10    "*"     5    ' 

^     2x—l     x-^7  .  x  —  4:     x-S 
^'    — t; ;:; i A t: — 


7x-6     Sx  +  2  ,  x  +  1     5a;-10 

8  3  4  12 

2a;  +  3  ,  x-2      5a7  +  4      2a;  — 4 

9  6  12  3      ' 

g     2a7  +  3  ■  x  +  S      182:  +  5     x-S 
2x  4a;  Sa;'-^  x 

X      2a:— 11      a;  +  3  ■  x—7      x~l 
*    2  3  4     "^     6  12  * 

4aV    a-^b  .  Ah'     a'b  +  ab' —  4:a^ 
'     b'         ab    ~^  a'  "^  a'b' 

-     5a;-ll      a;-l  ,  lla;-l      12a;-5 

O. ~77^ T 


4  10  12  8 

Q     a;  +  l  ,  3a;-4  ,  1      6a7+7 

^'  'y""^~5^"^4     8  • 


130  SCHOOL   ALGEBRA. 


10     2a?-6      8^-4  .  56a7-48 


11. 


bx  15a;  45a; 

lla;y  +  ^      5.v'  — 3      6x'  —  b 
a;y  X7/^  x^y 


12.      ^      I      1 1     J  2a;  — 2  J  y-6z_ 

2x^y      Qy'^z     2xz^       4:x'^z^        4:x'^yz 

162.   When  tlie  denominators  have  compound  expressions. 

(1)  Simplify   4 — -• 

^  ^         ^    ^      a-b        a  +  b       o?-b^ 

TheL.C.D.  is(a-5)(a  +  6). 

The  multipliers  are  a  +  6,  a  — 6,  and  1,  respectively. 

(a  +  6)  (2  a  +  6)  =      2  a^  +  3  a6  +  J^  =  1st  numerator. 

—  {a  —  h){2a  —  h)  =  —  2a?  -{■  2,ah  —  V^  =  2di  numerator. 

-I(6a6)  =  —Qah  =  3d  numerator. 

0  =  sum  of  numerators. 

.'.  sum  of  fractions  =  0. 

(2)  Simplify   ^  +  ^  +  ^-3 


r-4 

The  L.  e.  D.  is  {x  -  2)  {x  -2,){x-  4). 
(a;  -  1)  (a;  -  3)  (re  -  4)  =    a;^  -    8  a;^  -h  19  rs  -  12  =  Ist  numerator, 
(a;  _  2)  (a;  -  2)  (ar  -  4)  =    3?-   8  a;^  +  20  a;  -  16  =  2d  numerator, 
(a;  -  2)  (a;  -  3)  (a;  -  3)  -    a^  -    8  a;^  +  21  a;  -  18  =  3d  numerator. 


3a;3- 

24a;2-f 

■  60  a;  —  46  =  sum  of  numera 

.*.  sum  of  fractions 

3a;3 

-24aj2 
-2){x- 

+  60  a; -46 
-3)(a;-4) 

Exercise  48. 

Simplify  : 

1       ,       1 

'•r^- 

2 

a;  +  6     x-b 

1-a;' 

1            1 

1.      ^  +  ^      ■ 

^-y' 

2. 

l+a;      1— a;  '   x  —  y      {x  —  yY 


FRACTIONS.  131 

6.    37  — y      {x-yy  ^     x-\-y     x-y        ^xy 

x-\-y      (x  +  yY  ^  '   x  —  y     x-j-y  x^  —  y' 

6.    ^ 1 ^ 10.    ~ h     ^      •      ^^^ 


2a(a  +  5)      2a(a— 5)  a—x     a-\-x     c^-{-x^ 

7.        1+a;              \-x  ^^  1         2      J       1 

'    \-{-x-\-a?      \  —  x^x^  '  X     x-}-l     x-\-2 

g         1          (a-Scf  ^2     6-2a 2 1_ 

a-3c     a^-27/  "  9-a^      3+a     3-a 

13.    — V--i 


14. 
15. 
16. 


x-{'2y     x  —  2y     x"^  —  Ay^ 
_1 2     _J 1_ 

y-1    y    2/  +  1    2/'-i 
-^ 1 L_  +  i. 

07—1  37^  —  37        X 

b  ah  ah^ 


a-^b      {a^-hy     ia-\-bf 

17.    37 +  y        2^7        x''{x  —  y) 
y        x-\-y     yix'  —  y'') 
Hint.   Reduce  the  last  fraction  to  lowest  terms. 
3      .       4a  5a^ 


18. 
19. 


x  —  a     {x  —  of     (07  —  obf 
1        ,        a^  07  +  <2 


07-2a     ot'  — 8a'     07' +  2  ao;  +  4  a' 


„       07^-207  +  3    ,  07-2  1 


07^+1  07^  — 07+1        07+1 

21  1  I  37—1  ■    07^  +  07—3 
'     07-3'^O7^+3o7  +  9"^      07»-27     * 

22  07^  +  807+15     07-1 

'     07^  +  707+10        07-2 

Hint.   Reduce  the  first  fraction  to  lowest  terms. 


132  SCHOOL   ALGEBRA. 

23      ar^  —  6 ax -\- 6 a^      x—7a 
x"^  —  8ax-\-15a^     x  —  ba 

24.    -i.4  ' 


x-2     x''  —  2>x-\-2     x'-4:x-^Z 

Hint.    Express   the   denominators   of  the  last  two  fractions   in 
prime  factors. 

1.2  3 


25. 
26. 
27. 


a'-7a  +  12     a^-4a4-3     a^-5a  +  4 

3 4 

lOa'^+a-3     2a^  +  7a-4' 

3 1 

2-x-Qx^      l-x~2x' 


163.  Since  -j-  =  a,  and  — -  =  a,  it  follows  that 

0  ~  0 

The  value  of  a  fraction  is  not  altered  if  the  signs  of  the 
numerator  and  denominator  are  both  changed. 

It  follows,  also,  by  the  Law  of  Signs,  that 

The  value  of  a  fraction  is  not  altered  if  the  signs  of  any 
even  number  of  factors  in  the  numerator  and  denominator 
of  a  fraction  are  changed. 

164.  Since  changing  the  sign  before  a  fraction  is  equiva- 
lent to  changing  the  sign  before  the  numerator  or  the 
denominator,  it  follows  that 

The  sign  before  the  denominator  m,ay  be  changed,  "provided 
the  sign  before  the  fraction  is  changed. 

Note.  If  the  denominator  is  a  compound  expression,  the  beginner 
must  remember  that  the  sign  of  the  denominator  is  changed  by 
changing  the  sign  of  every  term  of  the  denominator.     Thus, 


These  principles  enable  us  to  change  the  signs  of  frac- 
tions, if  necessary,  so  that  their  denominators  shall  be 
arranged  in  the  same  w'der. 


FRACTIONS.  133 

(1)  Simplify  l-^,  +  l^- 

Changing  the  signs  before  the  terms  of  the  denominator  of  the 
third  fraction  and  the  sign  before  the  fraction,  we  have 
2  3  2x-3 

X      2a;  —  1      4a;'^  —  1 
The  L.G.'D.='x{2x-  l)(2a;  +  1). 

2(2a;-l)(2x  +  l)=      8^2_2    =  Ist  numerator. 

—  Sx{2x  +  1)  —  —  6x^  —  3x  =  2d  numerator. 

—  a; (2 a;  — 3)  =  ~  2a;^  +  3  a;  =  3d  numerator. 


sum  of  the  fractions  =  — 


—  2    =  sum  of  numerators. 
2 


a;(2a;-l)(2a;  +  l) 
(2)  Simplify 

I + 1 + 1 

a{a~b){a  —  c)      b(b-~  a)  (b  —  c)      c{c  —  a)  (e  —  b) 

Note.  Change  the  sign  of  the  factor  {h  —  a)  in  the  denominator 
of  the  second  fraction,  and  change  the  sign  before  the  fraction. 

Change  the  signs  of  the  two  factors  (c  —  a)  and  (c  —  h)  in  the  de- 
nominator of  the  third  fraction.     "We  now  have 

I 1 + 1 

a{a  —  h){a  —  c)      b{a  —  b)  {b  —  c)      c{a—  c)  {b  —  c) 
The  L. C. D.  =  abc {a  -b){a-  c) {b  -  c). 

he  (b  —  c)  =•  b'^c  —  hc^  =  1st  numerator. 

—  ac(a  —  c)  =  —  o?c  +  ac^  =  2d  numerator. 

ab  {a  —  b)  =  a^b  —  ab^  =  3d  numerator. 

a^b  —  a^G  —  ab^  +  ac^  +  b'^c  —  bc^  =  sum  of  numerators. 

.  a2(&  -  c)  -  a(6»  -  c2)  +  bc(b-  c), 

-[a2-a(6  +  c)  +  &c][6-c], 

=■  [a^  -~ab  —  ac  +  bc][b  —  c], 

=  [(a2  -  ac)  -  (ab  -  be)][b  -  c], 

=-[a{a  —  c)  —  b{a  —  c)]  [b  —  c], 

=  {a-b){a-c){b-c). 

.-.  sum  of  the  fractions  =  Ja-h){a- r){b  -  c)    ^    1 
abc  (a  —  o)  (a  —  c)  (o  —  c)     abo 


134  SCHOOL   ALGEBRA. 


Exercise  49. 
Simplify ; 

^  3j  I  JO  JO 

-        a      .     2ta     .     2ax 

/5. {-• 1~ 


a  —  x  '  a-^x  '  x^  —  0^ 
•3  2        .        15 


2a  — 3      3  +  2a     9  -  ^a^ 

h         a  —  h      ¥  —  a^b 

^3,5  2x-1 

5.-4- 


X        1-2X        4:X'  —  1 

1.1,1 


{x-\-ay      (a  —  xf     x^  —  (^ 

7.       1       I         x  —  y  xy  —  ^x^ 

x  —  y     x^ -{■  xy  +  y''       y^  —  a? 


(a;-2)(a;-3)      (:r-l)(3-^)      (a;-l)(a7-2) 
-.  5c  ,  a^  ,  a5 

^'     7 TT TT+T TTTl T    r 


(c  —  a)  (a  —  6)      (a  —  b){b  —  c)      {b  —  c){c  —  a) 
10.  ^  +  g  [  <^  +  g  [  «  +  5 


11. 


(a-5)(a-c)      (^)-c)(5-a)      (o-a)(c-b) 
3  4  6 


(a  —  b){b--  c)      {b  —  a)(c  —  a)      (a  —  c)  (c  ~  b) 


12.    1 + 1 i 

x{x-y){x-z)     y{y-x){y-z)     xyz 

13  o^  —  be         .         5^  +  ag         [         <?*  +  Q^^ 

{a-b){a-cy  {b-a){b-\-c)      {c-a){c-\-b) 


FRACTIONS.  135 


Multiplication  and  Division  of  Fractions. 
165.   Multiplication  of  Practions. 
Find  the  product  of  -  X  -3" 

Let  7  =  ^7  and  -,  =  ?/. 

Then  a = 6ir,  and  c  =  dy. 

The  product  of  these  two  equations  is 

ac  =  hdxy. 

ac 


Divide  by 

hd, 

But 

a      c 

Therefore, 

a      c       ac 
h      d~hd 

To  find  the  product  of  two  fractions,  therefore, 

Find  the  product  of  the  numerators  for  the  required 
numerator,  and  the  product  of  the  denominators  for  the 
required  denominator. 

166.    In  like  manner, 

h      d    f~hd    f~bdf 
and  so  on  for  any  number  of  fractions. 

Agam,  y=-x-  =  -. 

In  like  manner, 
^d 


167.  Division  of  Fractions.  If  the  product  of  two  numbers 
is  equal  to  1,  each  of  the  numbers  is  called  the  reciprocal  of 
the  other. 


136  SCHOOL    ALGEBRA. 

The  reciprocal  of  -  is  -» 
ha 

r.  h  ^,  a     ha      -, 

for  _X-  =  — =1. 

a     o      ab 

The  reciprocal  of  a  fraction,  therefore,  is  the  fraction 
inverted. 

Since  7-t--  =  1, 

0      b 

and  -  X  7  =  1,  it  follows  that 

a      h 

To  divide  hy  a  fraction  is  the  same  as  to  multiply  hy  its 

reciprocal. 

To  divide  hy  a  fraotioii,  therefore, 

Invert  the  divisor  and  multiply. 

Note.  Every  mixed  expression  should  first  be  reduced  to  a  frac- 
tion, and  every  integral  expression  should  be  written  as  a  fraction 
having  1  for  the  denominator.  If  a  factor  is  common  to  a  numera- 
tor and  a  denominator,  it  should  be  cancelled,  as  the  cancelling  of  a 
common  factor  hefore  the  multiplication  is  evidently  equivalent  to 
cancelling  it  after  the  multiplication. 

(1)  Find  the  product  of  |^X  |^X  -^. 
^  ^  ^  2>cd'     bab      8a'c'd^ 

2a?h     QcH      baVc  _2x  Qx  5a^¥(^d _  b^ 

?>cd^     bah      8ahW     Sx5x8a^c^d*^     2d^'        \ 

(2)  Find  the  product  of 

^'  -  .y'         X  ^.y  -  2.v'  ^  x'-xy 
x^—Sxy-{-2y'^       x^  +  xy       (x  —  yY 

x^  —  y^  xy  —  2  y^      x^  —  xy 

x^  —  3xy  +  2y^       x^  -j-  xy       (x  —  yY 

_  ix-y)(x  +  y)  ^^y(x-2y),,      x(x-y) 
(x-y){x-2y)       x{x  +  y)       {x-y){x-y) 

y 

^  X  -y 


FRACTIONS.  137 

Note,   The  common  factors  cancelled  are  (a;  —  3/),  {x  +  y),  (05  —  2y), 
X,  and  x  —  y. 

(3)  Find  the  quotient  of       "^  ^^ 


(a  —  xy      a*  —  x"^ 

ax  ah     ^  ax  {a  —  x){a  +  x) 

(a  —  o;)'^     a^  —  x^     (a  —  x)  {a  —  x)  ab 

^  x{a  +  x)^ 
h{a  —  x) 

The  common  factors  cancelled  are  a  and  a~x. 

Exercise  50. 

Simplify  : 

1.    12^x^-4^.  10. 

If      9x'z'  ' 

v/-^     3^6V^20mV  ,, 

^*   ~A ^  TTi — n"'  ■*•■*■• 

7mx2/       3aV'  '    a(a  +  5)      a(a'^—b') 

8a;^3/^      15  WW  a'4-4aa;      ax-{-4:X^ 

g     16a^_^4a^^  ^^      a;'-y*   .  x'  +  xy 

21a^7/'^  '  Sx^y  '    {x  —  yy       x  —  y 

^6     Z^-^M^  15      ar'  +  a'^       a;  +  3 « 

12z'  '  862//  '   a:^-9a'       a;  +  a  ' 


8  a' 

4:  a' 

a«-5'' 

a'  +  ab  +  b' 

^^  +  ,V^  . 

Sx'  +  Sy' 
x-\-y 

ah-h' 

.          h^ 

x'  +  y'     x-\-y  .  a^-^b^'  a^-ab-\-  b^ 

8.    i^!zLEx— ^^— .  17.      ^'-^     X    ^'~^^ 


a  2a;' -4a;  x'-4:X-b     a;2_^2a;-3 

9.        ^^      x^^^.  18.    (l-A^(^±I^. 

5a:- 10        4a;'  V       ^V      \    ^V    J 


K 


138  SCHOOL   ALGEBRA. 


20.    /8^_iy     4^-  +  2^y \ 


21. 


22.    8a^x— X— X  ^^^-^^' 


c         8a^       ^c^       4:(b^-bd) 


z^y-g^ 


24.  r(a+^')^-^^  ^  (a  +  cy-hn^Aa 

\^a^ -\- ab  —  ac  a  J     ab  —  b'^  —  bc 

{a  +  bY~c\   c^-ja-bY        c-a-b 
a'-(b-cy     c'-(a  +  by     ac-a'  +  ab 

26     (^  ~  Q^)'  ~  ^'  v^  x^  —  (b  —  ay      ax-\-  0^  —  db 
(x-by-a'     x'-(a-by     bx-ab  +  b^' 

_^     a''-2ab  +  b'-c'a-j-b-c 


L  x{x+l)  x'  +  x     J  a? 

29        2  gar^  +  2  a^a:  x^  —  a^        x-\-a 

'    {x-ay{x-^ay      2{x^^a^)        ax   ' 

30.    a^-h^-f-^lOy^U ?f-\ 

a^  —  ab  —  ac         \       a  +  b-\-cJ 

a^^a^b'-\-b'      a  +  b  ya'-h' 
a^-b'  a^j^b'  a 

32     c^^-1xy-{-\2y\^^^   x'-{-xy-2y' 
a:'  +  5a:y  +  6y'      x^ -\- 2> xy  —  ^y^ 


FRACTIONS.  ^  139 

168.  Complex  Fractions.  A  complex  fraction  is  one  that 
has  a  fraction  in  the  numerator,  or  in  the  denominator,  or 
in  both. 

(1)  Simplify     ^^ 


2>x  ^oc         2>x     A:x  —  1      3a: 


X  - 


x-\     Ax -I       14  1       4a;— 1 

4 

_    12a? 

407-1* 

Note.  Generally,  the  shortcvst  way  to  simplify  a  complex  fraction 
is  to  multiply  both  terms  of  the  fraction  by  the  L.  C.  D.  of  the  frac- 
tions contained  in  the  numerator  and  denominator.     Thus,  in  (1),  if 

1  O/v. 

we  multiply  both  terms  by  4,  we  have  at  once 

4a;— 1-        ^ 

a-\-x      a  —  X 

(2)  Simplify    l~l^l:^l-      .i_gU,^   K   /^ 

a  —  X     a-\-  X 

The  L.  0.  D.  of  the  fractions  in  the  numerator  and  denom- 
inator is 

(a  -~x)(a-{-  x). 

Multiply  by  {a  --x){a-\-x),  and  the  result  is, 

(g  +  xy  -  (g  -  xY 

{a  +  xf  +  la-xy 
___{a?  +  2a:Jc  +  x"")  -  (g^  -  2 ga:  +  oi^) 

(a'  +  2ga;  +  ^')  +  («'  -  2ga;  +  x') 
_:a^  +  2ax-hx'-a^  +  2ax-x' 

a^  -}-  2ax  -}-  x^  -{-  a^  —  2ax  -\-  3^ 

4ga: 

'~2g'^  +  2a:' 
__   2ga7 

a'  +  a^' 


140  SCHOOL   ALGEBRA 

(3)  Simplify  - 


1  +  ^+         "^ 


l-x-\-x^ 


1- 


x{l  —  X  -\-  x'^) 


I  \  X  \  ^  {l  +  x)(l-~x-{-x')+x 

l—x-\-x^ 


X^  -f-  3? 


l-Yx-^x" 
x{\-\-x^x'') 


\-^X^X^-{x  —  X^-^3?) 

~    1  +  ^'   ' 

Note.  In  a  fraction  of  this  kind,  called  a  continued  fraction,  we 
begin  at  the  bottom,  and  reduce  step  by  step.     Thus,  in  the  last 

example,  we  take  out  the  fraction ,  and  multiply 

1~  X  -It  X^ 

the  numerator  and  denominator  by  1—x  +  x^,  getting  for  the  result, 
x{l-x  +  x'^)  ^^ .^1^  simplified  is  ? " ^'  +  ^' 


{l+x){l—x-\-x^)+x  l+a;  +  a^ 

Putting  this  fraction  in  the  given  complex  fraction  for 

X 


1  +  a;  +  -- 


\  —  X  ■{■  x^ 

we  have  '■ -• 

1      x  —  x^-\-cir 

1  +a;  +  ar' 

Multiplying  both  terms  by  1  +  a;  +  re',  we  get 

x{\  +  a;  +  a^) 
l-^x  +  a^  —  x  +  x^  —  s? 

X  -V  x^  ->t  X* 
l+x"     ' 


Simplify  : 


X       y 
-  +  - 

z 
m 


2.  — i. 

y 

z 

X 


^-dd 


3.    ^ 


mn 


FRACTIONS.  141 


Exercise  51. 


m 


m  + w 


8. 
9. 

2m-\-n      -^ 
m-\-n 

1     » 

^  +  ,v' 

.-r'^  —  ^y  4-  y' 

x-y 
1              a 

n 

«_^      a^-b' 

a                b 

d  ab-\-h^     a^-\-  ab 

a;  —  i  ab      ac      DC 


'    ^ L_  a'-jb  +  cf 

x-}-l  ab 

5.   ^-y.  12.  ^-^. 

a;  +  2/  oca 


1        1        1 


6. 13. 


7.  _3__  j£:^.      14.  1 


a5" 

a(? 

^»(? 

a^- 

-(b- 

-^)'^ 

1    1 

a 

X 

J-  1 

1+^ 

v  +  - 

2x' 

142  SCHOOL    ALGEBRA. 

x-\-y 


15.    1 16. 


1  1  ,        >  1 


1+— 1--  ^-2/  + 


17.    —        ^ 


^  [      1        y{xyz-\-x-\-z) 


hG-\-  ac-\-  ah 


19. 


a  — 

1 

1  ^ 

— 

1  , 

^-1 

a 

T^ 

6 

i^ 

c 

:«^ 

1 
l 

m'  +  n^ 

20.  A  -  y    ^  -  y'^  V.  /^  +  y  ,  ^'  +  y'\ 


1  ,       1 


Exercise  52. 

Examples  for  Review. 

1.  Find  the  value  of  Va'  +  6^  +  c^  -  (a  -  ^>  -  c)^  when 

a  =  2,  5  =  —  2,  and  ^  =  4. 

2.  Reduce  to  lowest  terms  — -^ ^  ~~ — • 

3a;'  +  13:r"'  +  17^  +  6 

Simplify  : 

1  a;-4  ,  x-Z 


{x--2>){x-2)      {x-l){x-^)      {x  —  \){x-2) 


FRACTIONS.  143 


4.       ^'  +  yg       [       y'  +  ^2      J       z'  +  ^y 

{x-y){x-z)      {y-z){y~x)      {z-x){z-y) 

6.    Z^-^  +  lzi^V^-^ i^^\ 

\l-\-x  X     J   '   \l  +  X  X    ) 

6.  Y^+_^^ 3 

c\x—c     x-\-2cJ     x^  -\-  ex  —  2(? 

xy    V^''-2/'  ^'  +  2/7 

x^-x-2  x''  +  x-2        0:^-4* 

9.    f^±l^  .  fV  ff  .  2__^\ 
\x-\-y    '  y)      \y  '  x  +  yj 

10.  fi_^^+JL2_yi+.^__ii.\ 

V       a;-l^a7-3yV  ^4-1     ^  +  3y 

11.   x'-xy  +  y'^^o^  —  y'  .  (y-a;^     ' 
x'  +  xy  +  y'     x'  +  f'  {x-\-yy 

2a  —  h  —  c      .       2h  —  c  —  a      ,       2c  — a  — J 
1^.    _ ^ r~r71 TTt T~r 


(a  —  b)(a  —  c)      (b  —  c)(b  —  a)      (c  —  a){c  —  b) 
1  2  1,1      1,1 


^^     x+l      {x  +  2){x+l)  a'      b'      a      b 

a;  +  2      {x  +  l){x-\-2)  a"     b""      a     b 

1R     A      1-^  ,  l  +  2^^/a;+l\ 

''•  V     iT^+ir^j(,27+i/ 
16.  ^-^y"x  ^'-^y  +  .v'  v^^(^'-y') 

a:(a:  — y)      a;^4-2a:y  +  43/^         ^'  +  2/^ 

17     2.2.2     Z>  +  g  —  g      c  +  g  —  &      a-{-b  —  c 
a     b      c  be  ac  ab 


Co 

CHAPTER  X. 
FRACTIONAL    EQUATIONS. 
169.   To  reduce  Equations  containing  Tractions. 

(1)  Solve  |-^  =  y-9. 

Multiply  by  33,  the  L.C.M.  of  the  denominators. 
Then,  11a;  -  3.^  +  3  =  33a;- 297,  ■ 

11  a;  -  3  a;  -  33  a;  =  -  297  -  3, 
-25  a;  =  -300. 
.-.  a;  =  12. 

Note.  Since  the  minus  sign  precedes  the  second  fraction,  in  remov- 
ing the  denominator,  the  +  (understood)  before  x,  the  first  term  of  the 
numerator,  is  changed  to  — ,  and  the  —  before  1,  the  second  term  of 
the  numerator,  is  changed  to  +. 

Therefore,  to  clear  an  equation  of  fractions, 

Multiply  each  term  hy  the  L.  0.  M.  of  the  denominators. 

If  a  fraction  is  preceded  by  a  minus  sign,  the  sign  of 
every  term  of  the  numerator  must  be  changed  when  the 
denovfiinator  is  removed. 

(2)  Solve  ^^-^^  =  ^-=^-^^:i^. 
^  x  —  b     x  —  %     x  —  d>     x—^ 

Note.  The  solution  of  this  and  similar  problems  will  be  much 
easier  by  combining  the  fractions  on  the  left  side  and  the  fractions 
on  the  right  side  than  by  the  rule  given  above. 

{x  -  4)  (a;  -  6)  -  (a;  -  S)'^      (a;  -  7)  (a;  -  9)  -  (a;  -  8)' 
(a;  -  5)  (a;  -  6)  {x  -  8)  (a;  -  9) 


FRACTIONAL  EQUATIONS.  145 


By  simplifying  the  numerators,  we  have 
-1  -1 


(a;  -5){x-  6)     {x  -  8)  (a;  -  9) 
Since  the  numerators  are  equal,  the  denominators  are  equal. 
Hence,  {x  -5){x-6)  =  {x~  8) {x  -  9). 

Solving,  we  have  x  =  7. 


Exercise  53. 

Solve : 

1     3^-1      2a;+l      4a;-5_. 
4  3.5 

6  4 

g     5x+l      19:r4-7      3a7-l_7a:-l 
3  9  2  6     ' 

.      .   .  X      Sx-2      lla:  +  2      2-7:?; 
7  2  14  3 


5.   — 2  = \-x. 


g     63;+7  ■  7a7-13_2a;  +  4 

9      "^      27  3      *         ■ 

^     9x  +  5     8x~7__S6x+lb     41 
14  14  56  56* 

8  ^  +  3      .r-2_3a;-5  .  1 

2  3  12         4* 

9  n      /^3a:-l      2:r+l\      1Q      /2;r-5      7a;-l 

V4"^3y  V3'^8 

10     "^-^-4  ■  3.r-l      5(a;-l)_3(3a;-l)  ■  x 
9      "^      5  6  20       "^7 


146  SCHOOL    ALGEBRA. 

,,     ^        27^-1      2(4^-l)_9;r-5      llx-2,cyo 
n.    6^  -  -        _-^  3""^^^' 

,^     10a;+ll      12a;-]3      ._7-6a; 

12.  — ^  -  4--^-. 

13.  B_^        5_^_^        .  1 


y-4      2(y-4)      2(y-4)      2 
2  5  8  a: 


Lt. 

07-1      2(a:-l)      3(:r-l)      a;-l  '  18 

1^. 

2a:-3      g_  07  +  5       11 
2a;-4             3a;-6       2 

16. 

10-7:^_5a:-4      ^^     2r?;  + 1           8       _2a;-l 
6  — 7^7          bx         ''  '    2x-l     4^7^-1      2^7+1 

17. 

5  +  8^7     45-8^              5-207     2-7o7      5a7''  +  4 

3  +  207         13-207         "    '       07-1            07+1            07^-1 

18. 

5o:-l_5o7-3        „„        6         07+2         07'^     _q 
2o7  +  3      2o7-3        ""'    07  +  2      07-2  '07^-4 

19. 

X     07^-507     2                     4         07+1     ^'--3_Q 
3      3o:-7      3          "  •    1  +  07      1-07     l-07»        • 

*?4 

2o7+l      7o7~l      207^-307-45 

3o7-3      607  +  6           4o7^-4 

25. 

^^-^+l+^^  +  ^+l-.2o7. 

07—  1                     07+1 

26. 

9o7  +  5       3o7^-51o7-71_15o7— 7 

6(07-1)    '            18(07-- 1)                9(07+1) 

27. 

4+7                37         _^ 

07  +  2        07  +  3        07^+507+6 

28. 

1111 

FRACTIONAL  EQUATIONS.  147 

170.  If  the  denominators  contain  both  simple  and  com- 
pound expressions,  it  is  best  to  remove  the  simple  expressions 
first,  and  then  each  compound  expression  in  turn.  After 
each  multiplication  the  result  should  be  reduced  to  the 
simplest  form. 

(1)  Solve  ^^+I^  =  i£±-^. 

Multiply  both  sides  by  14. 

Then,  8x  +  5  +  1^^l-^^^=^  =  8x  +  12. 

3x4-1 

Transpose  and  combine,      ^  ^~  ^  =  7. 
^  3a; +  1 

Divide  by  7  and  multiply  by  3  a;  +  1, 

7a;- 3  =  3a; +  1. 

.•.a;  =  l. 

(2)  Solve -  =  ^ ^— _. 

^  ^  4  4  10 

Simplify  the  complex  fractions  by  multiplying  both  terms  of  each 
fraction  by  9. 

Then.  2I^i£.i_I^ZL27. 

36  4  90 

Multiply  both  sides  by  180. 

135 -20a;  =  45 -14a; +  54, 
-6a;  =  -36. 
.•.x  =  G. 


Solve 


1. 


2. 


Exercise  54. 
4a; +  3      2x-5_2x-l      y  -c,^ 


10         5a;- 1  5 

9a7  +  20^4a;-12  .  x 
36  5a;-4'^4' 


X 


148  SCHOOL    ALGEBRA. 

18  13a;- 16  9 

^     6a;+7  .  7x-  13^2:r  +  4 

9  6^  +  3  3 

6^+7  2^-2_2:r  +  l 

*        15  7a; -6  5      * 

6^7+1  2.r-  4  ^2a;  — 1 

15  To;- 16  5 

lla;-13      22a;-75_  13a;+7 
14  28  2  (3  a; +  7)* 

2x-{-8^      13.r-2      x_7x     07+16 

9  17a:-32'^3      12  36 

l  +  3a:     2x  —  2l 


7. 


9. 


15         14  (a; -1)         21  6  105 


2a;  — 5,     a;  — 3    _4a;  — 3      -,  ^ 

'^*  ~T~  +  2^rri5-~To~~^T^' 

171.  Literal  Equations.  Literal  equations  are  equations 
in  whicli  some  or  all  of  the  given  numbers  are  represented 
by  letters;  the  numbers  regarded  as  known  numbers  are 
usually  represented  by  the  j?rs^  letters  of  the  alphabet. 

(1)    (a  —  x)  (a  -t  x)  ==  2a^  -j-  2 ax  —  x^. 
Then,  a^  -  x"^  =  2a^  +  2ax-x\ 

—  2ax  =  a?. 

'  :.x=^-^. 

2 

Exercise  55. 
Solve : 

1.  ax-\-2b  =  ?>hx-{-4:a.    3.  {a-\-x)(h  +  x)  =  x{x  -  c), 

2.  x'-\-b''  =  {a~x)\         4.  (x-a)(x+h)  =  {x-h){x-c). 


5. 


FRACTIONAL  EQUATIONS.  149 

h 


- «,     x  —  a  _  /2  a;  —  Q^Y 
'    x-h~\^x-h) 

6.      g-t-Q^    _  'm-x  ^^     x-a_{x-hj 

'2         2a;-a* 


8. 


a  -f  ao; 

c-\-  ex 

c  +  c^ 

m  —  X 

ab-\-  hx 

an-j-nx 

x  +  2 

m-\-n 

x-2 

m  —  n 

m-\-n 

_m  —  n 

19. 


a 


10. 


I  b 

2-\-x       2  —  X  m 

a  +  hx_c-\-dx  ax-h  ax-^b 

a-^b        c-^d  X               X 

20.  = 

^x-\-a_?>x-b  CL              b 

4:r  +  6      2x-a  ^               a 


n     x.x.x 7  ex  -\-  d 

'    a     b      c        '  a     _2c? 


-  „     ax  —  b      bx-^-  c        7 

12. ^ —  =  abc. 

c  a 


ex 
d 


^ax        r.  ^  a-\~  X  —  o 

13.    ±^-Sa^8x.  -^~ , 

a  —  b  oo  ^  0 


22. 


-^     __^ 5(1    _   2bx  Q 

a  —  b      a-\-b      a^  —  b'^ 


c 


15. 


1   ,  _n__^jhn  23.     ^~^  +  ^  +  ^^0. 


16     ^  I      ^     .^     <^  24     ^+<^     g;— 5_2(a4-Z>) 

a      b  —  a      b-}-a  x  —  a     x-{-b  x 

25.    ^+^_a;-^»-^^^4-^^=^. 


„»     20a  — bx  .  9(? — a:r  ,  &d—cx     i^ 

-ib.  — 1 1 — — - — ==ia 

5a  3(?  2c? 

27        «^  ^   ~  ^    I     ^  (^  ~  ^)  ::^  ^ 

6         2^    "*"       3c^ 


150  SCHOOL    ALGEBRA. 

172.    Problems  involving  Praotional  Equations. 

Ex.  The  sum  of  the  third  and  fourth  parts  of  a  certain 
number  exceeds  3  times  the  difference  of  the  fifth  and  sixth 
parts  by  29.     Find  the  number. 

Let  X  =  the  number. 

Then     -  +  -  =  the  sum  of  its  third  and  fourth  parts, 

^  _  2  =  the  difference  of  its  fifth  and  sixth  parts, 
5      6 

3|-  —  ^j=3  times  the  difference  of  its  fifth  and  sixth  parts, 

_-l c5( \=  the  given  excess. 

3     4        \5     6/ 

But  29  =  the  given  excess. 

3     4        V^     6/ 

Multiply  by  60  the  L.  C.  D.  of  the  fractions. 
20  a;  +  15  a;-  36aj  +  30  a;  =  60  x  29. 
Combining,  29  a;  =  60  x  29. 

.-.  a;  =  60. 

Exercise  56. 

1.  The  sum  of  the  sixth  and  seventh  parts  of  a  number 
is  13.     Find  the  number. 

2.  The  sum  of  the  third,  fourth,  and  sixth  parts  of  a 
number  is  18.     Find  the  number. 

3.  The  difference  between  the  third  and  fifth  parts  of 
a  number  is  4.     Find  the  number. 

4.  The  sum  of  the  third,  fourth,  and  fifth  parts  of  a 
number  exceeds  the  half  of  the  number  by  17.  Find  the 
number. 

5.  There  are  two  consecutive  numbers,  x  and  rr-f-1, 
such  that  one-fourth  of  the  smaller  exceeds  one-ninth  of  the 
larger  by  11.     Find  the  numbers. 


FRACTIONAL  EQUATIONS.  151 

6.  Find  three  consecutive  numbers  such  that  if  they 
are  divided  by  7,  10,  and  17,  respectively,  the  sum  of  the 
quotients  will  be  15. 

7.  Find  a  number  such  that  the  sum  of  its  sixth  and 
ninth  parts  shall  exceed  the  difference  of  its  ninth  and 
twelfth  parts  by  9. 

8.  The  sum  of  two  numbers  is  91,  and  if  the  greater  is 
divided  by  the  less  the  quotient  is  2,  and  the  remainder 
is  1.     Find  the  numbers. 

Hint.   Dividend -Bemainder^Q^^^.^^^^ 
Divisor 

9.  The  difference  of  two  numbers  is  40,  and  if  the  greater 
is  divided  by  the  less  the  quotient  is  4,  and  the  remainder 
4.     Find  the  numbers. 

10.  Divide  the  number  240  into  two  parts  such  that  the 
smaller  part  is  contained  in  the  larger  part  5  times,  with  a 
remainder  of  6. 

11.  In  a  mixture  of  alcohol  and  water  the  alcohol  was 
24  gallons  more  than  half  the  mixture,  and  the  water  was 
4  gallons  less  than  one-fourth  the  mixture.  How  many 
gallons  were  there  of  each  ? 

12.  The  width  of  a  room  is  three-fourths  of  its  length. 
If  the  width  Was  4  feet  more  and  the  length  4  feet  less,  the 
room  would  be  square.     Find  its  dimensions. 

Ex.  Eight  years  ago  a  boy  was  one-fourth  as  old  as  he 
will  be  one  year  hence.     How  old  is  he  now  ? 

Let  X  =  the  number  of  years  old  he  is  now. 

Then  a;  ~  8  =  the  number  of  years  old  he  was  eight  years  ago, 
and  X  +  1  ="  the  number  of  years  old  he  will  be  one  year  hence. 

...  a;-8  =  H^  +  l)- 
Solving,  a;  =  11. 


152  SCHOOL   ALGEBRA. 

13.  A  is  60  years  old,  and  B  is  three-fourths  as  old. 
How  many  years  since  B  was  just  one-half  as  old  as  A  ? 

14.  A  father  is  50  years  old,  and  his  son  is  half  that 
age.  How  many  years  ago  was  the  father  two  and  one- 
fourth  times  as  old  as  his  son  ? 

15.  A  father  is  40  years  old,  and  his  son  is  one-third 
that  age.  In  how  many  years  will  the  son  be  half  as  old 
as  his  father  ? 

Ex.  A  can  do  a  piece  of  work  in  6  days,  and  B  can  do 
it  in  7  days.  How  long  will  it  take  both  together  to  do 
the  work  ? 

Let        X  =  the  number  of  days  it  will  take  both  together. 

Then      -  =  the  part  both  together  can  do  in  one  day,  .     p| 

\  =  the  part  A  can  do  in  one  day,  ^ 

\  =  the  part  B  can  do  in  one  day, 
and      i  +  7  =  the  part  both  together  can  do  in  one  day. 

.•.1  +  1  =  1. 
6     7     a; 

7  a; -1- 6  a;  =  42. 

13  a;  =  42. 

a;  =  3^j. 

Therefore  they  together  can  do  the  work  in  3^  days. 

16.  A  can  do  a  piece  of  work  in  3  days,  B  in  4  days,  and 
0  in  6  days.  How  long  will  it  take  them  to  do  it  working 
together? 

17.  A  can  do  a  piece  of  work  in  2\  days,  B  in  3|-  days, 
and  G  in  4|  days.  How  long  will  it  take  them  to  do  it 
working  together  ? 

18.  A  and  Bean  separately  do  a  piece  of  work  in  12 
days  and  20  days,  and  with  the  help  of  0  they  can  do  it  in 
6  days.     How  long  would  it  take  0  to  do  the  work  ? 


FRACTIONAL  EQUATIONS.  153 

19.  A  and  B  together  can  do  a  piece  of  work  in  10  days, 
A  and  G  in  12  days,  and  A  by  himself  in  18  days.  How 
many  days  will  it  take  B  and  0  together  to  do  the  work  ? 
How  many  days  will  it  take  A,  B,  and  0  together  ? 

20.  A  and  B  can  do  a  piece  of  work  in  10  days,  A  and 
0  in  12  days,  B  and  0  in  15  days.  How  long  will  it  take 
them  to  do  the  work  if*  they  all  work  together  ? 

21.  A  cistern  can  be  filled  by  three  pipes  in  15,  20,  and 
30  minutes  respectively.  In  what  time  will  it  be  filled  if 
the  pipes  are  all  running  together  ? 

22.  A  cistern  can  be  filled  by  two  pipes  in  25  minutes 
and  30  minutes,  respectively,  and  emptied  by  a  third  in  20 
minutes.  In  what  time  will  it  be  filled  if  all  three  pipes 
are  running  together  ? 

23.  A  tank  can  be  filled  by  three  pipes  in  1  hour  and  20 
minutes,  2  hours  and  20  minutes,  and  3  hours  and  20  min- 
utes, respectively.  In  how  many  minutes  can  it  be  filled 
by  all  three  together  ? 

Ex.  A  courier  who  travels  6  miles  an  hour  is  followed, 
after  5  hours,  by  a  second  courier  who  travels  7|-  miles  an 
hour.  In  how  many  hours  will  the  second  courier  overtake 
the  first? 

Let  X  =  the  number  of  hours  the  first  travels. 

Then        a;  —  5  =  the  number  of  hours  the  second  travels,  ^ 
6  a;  =  the  number  of  miles  the  first  travels, 
and        {x  —  5)  7^  =  the  number  of  miles  the  second  travels. 
They  both  travel  the  same  distance. 

.-.    6a?=(a;-5)7J, 
or  12a;  =  15a; -75. 

.-.      a;  =  25. 

Therefore  the  second  courier  will  overtake  the  first  in  2(> 
hours. 


154  SCHOOL    ALGEBRA. 

24.  A  sets  out  and  travels  at  the  rate  of  9  miles  in  2 
hours.  Seven  hours  afterwards  B  sets  out  from  the  same 
place  and  travels  in  the  same  direction  at  the  rate  of  5 
miles  an  hour.     In  how  many  hours  will  B  overtake  A  ? 

25.  A  man  walks  to  the  top  of  a  mountain  at  the  rate  of 
22  miles  an  hour,  and  down  the  same  way  at  the  rate  of  4 
miles  an  hour,  and  is  out  5  hours.  How  far  is  it  to  the  top 
of  the  mountain  ? 

26.  In  going  from  Boston  to  Portland,  a  passenger  train, 
at  27  miles  an  hour,  occupies  2  hours  less  time  than  a  freight 
train  at  18  miles  an  hour.  Find  the  distance  from  Boston 
to  Portland. 

27.  A  person  has  6  hours  at  his  disposal.  How  far  may 
he  ride  at  6  miles  an  hour  so  as  to  return  in  time,  walking 
back  at  the  rate  of  3  miles  an  hour  ? 

28.  A  boy  starts  from  Exeter  and  walks  towards  Ando- 
ver  at  the  rate  of  3  miles  an  hour,  and  2  hours  later  another 
boys  starts  from  Andover  and  walks  towards  Exeter  at  the 
rate  of  2i  miles  an  hour.  The  distance  from  Exeter  to 
Andover  is  28  miles.    How  far  from  Exeter  will  they  meet? 

Ex.  A  hare  takes  5  leaps  while  a  greyhound  takes  3,  but 
1  of  the  greyhound's  leaps  is  equal  to  2  of  the  hare's.  The 
hare  has  a  start  of  50  of  her  own  leaps.  How  many  leaps 
must  the  greyhound  take  to  catch  her  ? 

Let  3  a;  =  the  number  of  leaps  the  greyhound  takes. 

Then       5  a;  =  the  number  of  leaps  the  hare  takes  in  the  same  time. 

Also,  let    a  =  the  number  of  feet  in  one  leap  of  the  hare. 

Then      2  a  =  the  number  of  feet  in  one  leap  of  the  hound. 

Hence  Sxx2a  or  6ax  =  the  whole  distance, 

and        (5  a;  +  50)  a  or  5ax  +  50  a  ==  the  whole  distance. 
.*.  6aa;  =  5aa;  +  50  a. 
.-.       a;  =  50, 
and  3  a;  =  150. 

Therefore  the  greyhound  must  take  150  leaps. 


FRACTIONAL  EQUATIONS.  155 

29.  A  hare  takes  7  leaps  while  a  dog  takes  5,  and  5  of 
the  dog's  leaps  are  equal  to  8  of  the  hare's.  The  hare  has 
a  start  of  50  of  her  own  leaps.  How  many  leaps  will  the 
hare  take  before  she  is  caught  ? 

30.  A  dog  makes  4  leaps  while  a  hare  makes  5,  but  3  of 
the  dog's  leaps  are  equal  to  4  of  the  hare's.  The  hare  has  a 
start  of  60  of  the  dog's  leaps.  How  many  leaps  will  each 
take  before  the  hare  is  caught  ? 

Note.  If  the  number  of  units  in  the  breadth  and  length  of  a 
rectangle  are  represented  by  x  and  a;  +  a,  respectively,  then  x{x  -\-  a) 
will  represent  the  number  of  units  of  area  in  the  rectangle. 

31.  A  rectangle  whose  length  is  2^  times  its  breadth 
would  have  its  area  increased  by  60  square  feet  if  its  length 
and  breadth  were  each  5  feet  more.     Find  its  dimensions. 

32.  A  rectangle  has  its  length  4  feet  longer  and  its  width 
3  feet  shorter  than  the  side  of  the  equivalent  square.  Find 
its  area. 

33.  The  width  of  a  rectangle  is  an  inch  more  than  half 
its  length,  and  if  a  strip  5  inches  wide  is  taken  oiFfrom  the 
four  sides,  the  area  of  the  strip  is  510  square  inches.  Find 
the  dimensions  of  the  rectangle. 

Note.  If  x  pounds  of  metal  lose  I  pound  when  weighed  in 
water,  1  pound  of  metal  will  lose  -  of  a  pound. 

X 

34.  If  1  pound  of  tin  loses  -^  of  a  pound,  and  1  pound 
of  lead  loses  -^  of  a  pound,  when  weighed  in  water,  how 
many  pounds  of  tin  and  of  lead  in  a  mass  of  60  pounds  that 
loses  7  pounds  when  weighed  in  water  ? 

35.  If  19  ounces  of  gold  lose  1  ounce,  and  10  ounces  of 
silver  lose  1  ounce,  when  weighed  in  water,  how  many 
ounces  of  gold  and  of  silver  in  a  mass  of  gold  and  silver 
weighing  530  ounces  in  air  and  495  ounces  in  water  ? 


156  SCHOOL   ALGEBRA. 

Ex.  Find  the  time  between  2  and  3  o'clock  when  the 
hands  of  a  clock  are  together. 

At  2  o'clock  the  hour-hand  is  10  minute-spaces  ahead  of  the 
minute-hand. 

Let  X  =  the  number  of  spaces  the  minute-hand  moves  over. 

Then  a;  —  10  =  the  number  of  spaces  the'hour-hand  moves  over. 
Now,  as  the  minute-hand  moves  12  times  as  fast  as  the  hour-hand, 
12  (a;  —  10)  =  the  number  of  spaces  the  minute-hand  moves  over. 
.-.  a;  =12  (a; -10), 
and  11a;  =  120. 

Therefore  the  time  is  lOfJ-  minutes  past  2  o'clock. 

36.  At  what  time  between  2  and  3  o'clock  are  the  hands 
of  a  watch  at  right  angles  ? 

37.  At  what  time  between  3  and  4  o'clock  are  the  hands 
of  a  watch  pointing  in  opposite  directions  ? 

38.  At  what  time  between  7  and  8  o'clock  are  the  hands 
of  a  watch  together  ? 

Ex.  A  merchant  adds  yearly  to  his  capital  one-third  of 
it,  but  takes  from  it,  at  the  end  of  each  year,  $5000  for 
expenses.  At  the  end  of  the  third  year,  after  deducting 
the  last  $5000,  he  has  twice  his  original  capital.  Hew 
much  had  he  at  first  ? 

Let  X  =  number  of  dollars  he  had  at  first. 

Then  i^-5000,  or  il^l^O, 

3  o 

will  stand  for  the  number  of  dollars  at  the  end  of  first  year, 
and  4/4.-15000\  ^^    16^-105000^ 

3V  3  /  9 

will  stand  for  the  number  of  dollars  at  the  end  of  second  year, 
^^^  4/16a;-105000\      ^OQQ    ^^   64a. -555000^ 

3\  9  /  27 

will  stand  for  the  number  of  dollars  at  the  end  of  third  year. 


FRACTIONAL  EQUATIONS.  157 

But  2x  stands  for  the  number  of  dollars  at  the  end  of  third  year. 
64  a; -555000 


27 
Whence  x  =  55,500. 


2x. 


39.  A  trader  adds  yearly  to  his  capital  one-fourth  of  it, 
but  takes  from  it,  at  the  end  of  each  year,  $800  for  ex- 
penses. At  the  end  of  the  third  year,  after  deducting  the 
last  $800,  he  has  1|-  times  his  original  capital.  How  much 
had  he  at  first  ? 

40.  A  trader  adds  yearly  to  his  capital  one-fifth  of  it, 
but  takes  from  it,  at  the  end  of  each  year,  $2500  for  ex- 
penses. At  the  end  of  the  third  year,  after  deducting 
the  last  $2500,  he  has  1-^  times  his  original  capital. 
Find  his  original  capital. 


41.  A's  age  now  is  two-fifths  of  B's.  Eight  years  ago  A's 
age  was  two-ninths  of  B's.     Find  their  ages. 

42.  A  had  five  times  as  much  money  as  B.  He  gave  B 
5  dollars,  and  then  had  only  twice  as  much  as  B.  How 
much  had  each  at  first  ? 

43.  At  what  time  between  12  and  1  o'clock  are  the  hour 
and  minute  hands  pointing  in  opposite  dir'ections  ? 

44.  Eleven-sixteenths  of  a  certain  principal  was  at  in- 
terest at  5  per  cent,  and  the  balance  at  4  per  cent.  The 
entire  income  was  $1500.     Find  the  principal. 

45.  A  train  which  travels  36  miles  an  hour  is  f  of  an 
hour  in  advance  of  a  second  train  which  travels  42  miles 
an  hour.  In  how  long  a  time  will  the  last  train  overtake 
the  first  ? 

46.  An  express  train  which  travels  40  miles  an  hour 
starts  from  a  certain  place  50  minutes  after  a  freight  train, 
and  overtakes  the  freight  train  in  2  hours  and  5  minutes. 
Find  the  rate  per  hour  of  the  freight  train. 


158  SCHOOL    ALGEBRA. 

47.  A  messenger  starts  to  carry  a  despatch,  and  5  hours 
after  a  second  messenger  sets  out  to  overtake  the  first  in  8 
hours.  In  order  to  do  this,  he  is  obliged  to  travel  2^  miles 
an  hour  more  than  the  first.  How  many  miles  an  hour 
does  the  first  travel  ? 

48.  The  fore  and  hind  wheels  of  a  carriage  are  respec- 
tively Qi"  feet  and  11-|  feet  in  circumference.  "What  distance 
will  the  carriage  have  made  when  one  of  the  fore  wheels 
has  made  160  revolutions  more  than  one  of  the  hind 
wheels  ? 

49.  When  a  certain  brigade  of  troops  is  formed  in  a 
solid  square  there  is  found  to  be  100  men  over ;  but  when 
formed  in  column  with  5  men  more  in  front  and  3  men  less 
in  depth  than  be^re,  the  column  needs  5  men  to  complete 
it.     Find  the  number  of  troops. 

50.  An  officer  can  form  his  men  in  a  hollow  square  14 
deep.  The  whole  number  of  men  is  3136.  Find  the  num- 
ber of  men  in  the  front  of  the  hollow  square. 

51.  A  trader  increases  his  capital  each  year  by  one- 
fourth  of  it,  and  at  the  end  of  each  year  takes  out  $2400 
for  expenses.  At  the  end  of  3  years,  after  deducting  the 
last  $2400,  he  finds  his  capital  to  be  $10,000.  Find  his 
original  capital. 

52.  A  and  B  together  can  do  a  piece  of  work  in  1-|-  days, 
A  and  0  together  in  IJ  days,  and  B  and  C  together  in  IJ 
days.  How  many  days  will  it  take  each  alone  to  do  the 
work  ? 

53.  A  fox  pursued  by  a  hound  has  a  start  of  100  of  her 
leaps.  The  fox  makes  3  leaps  while  the  hound  makes  2 ; 
but  3  leaps  of  the  hound  are  equivalent  to  5  of  the  fox. 
How  many  leaps  will  each  take  before  the  hound  catches 
the  fox? 


FRACTIONAL   EQUATIONS.  159 

173.  rormulas  and  Eules.  When  the  given  numbers  of  & 
problem  are  represented  by  letters,  the  result  obtained  from 
solving  the  problem  is  a  general  expression  which  includes 
all  problems  of  that  class.  Such  an  expression  is  called  a 
formula,  and  the  translation  of  this  formula  into  words  is 
called  a  rule. 

We  will  illustrate  by  examples  : 

(1)  The  sum  of  two  numbers  is  s,  and  their  difference  d] 
find  the  numbers. 


Let 

X  =  the  smaller  number 

then 

x  +  d=th.e  larger  number. 

Hence 

x  +  x  +  d=s, 

or 

2x  =  s-d. 

.•.^=-^. 

o«^ 

^ , ^     s-d+2d 

2 

s  +  <? 
^ ,  2 

Therefore  the  numbers  are     "t      and 


As  these  formulas  hold  true  whatever  numbers  s  and  d 
stand  for,  we  have  the  general  rule  for  finding  two  numbers 
when  their  sum  and  difference  are  given : 

Add  the  difference  to  the  sum  and  take  half  the  result  for 
the  greater  number. 

Subtract  the  difference  from  the  sum  and  take  half  the 
result  for  the  smaller  number. 

(2)  If  A  can  do  a  piece  of  work  in  a  days,  and  B  can 
do  the  same  work  in  h  days,  in  how  many  days  can  both 
together  do  it? 


160  SCHOOL   ALGEBRA. 

Let  X  =  the  required  number  of  days. 

Then,  -  =  the  part  both  together  can  do  in  one  day. 

X 

Now  -  =  the  part  A  can  do  in  one  day, 

a 

and  -  =  the  part  B  can  do  in  one  day ; 

b 

therefore        1  +  1  =  the  part  both  together  can  do  in  one  day. 
a     b 

•••-4" 

a     b     X 

ab 
Whence  x= — —r- 

a  +  b 

The  translation  of  this  formula  gives  the  following  rule 
for  finding  the  time  required  by  two  agents  together  to 
produce  a  given  result,  when  the  time  required  by  each 
agent  separately  is  known  : 

Divide  the  product  of  the  numbers  which  express  the  units 
of  time  required  by  each  to  do  the  work  by  'the  sum  of  these 
numbers ;  the  quotient  is  the  time  required  by  both  together. 

174.  Interest  Pormnlas.  The  elements  involved  in  com- 
putation of  interest  are  the  principal,  rate,  time,  interest, 
and  amount. 

Let   p  —  the  principal, 

r  =  the  interest  of  $  1  for  1  year,  at  the  given  rate, 

i  =  the  time  expressed  in  years, 

i  =  the  interest  for  the  given  time  and  rate, 

a  =  the  amount  (sum  of  principal  and  interest). 

175.  Given  the  Principal,  Eate,  and  Time ;  to  find  the  Interest. 

Since  r  is  the  interest  of  f  1  for  1  year,  pr  is  the  interest 
of  $p  for  1  year,  and  prt  is  the  interest  of  $p  for  t  years. 

.-.  i  =  prt.  (Formula  1.) 

Rule.    Find  the  product  of  the  principal,  rate,  and  time. 


FE ACTIONAL   EQUATIONS.  161 

176.  G-iyen  tlie  Interest,  Eate,  and  Time ;  to  find  the  Principal. 
By  Formula  1,  prt=^i. 

Divide  by  r^,  P^^^  (Formula  2.) 

177.  Given  the  Amount,  Eate,  and  Time ;  to  find  the  Principal. 

From  formula  2,        p  -\-prt  =  a, 
or  ^  (1  +  ^0  —  ^• 

Divide  by  1  +  r^,  p  =  --^ —  (Formula  3.) 

1  -f-  rt 

178.  Given  the  Amount,  Principal,  and  Eate ;  to  find  the  Time. 
From  formula  2,        p -\-  prt  =  a. 

Transpose  p,  prt  =  a—p. 

Divide  by  pr,  t  =    ~*'  (Formula  4.) 

pr 

179.  Given  the  Amount,  Principal,  and  Time ;  to  find  the  Eate. 
From  formula  2,        j9  -j-  prt  =  a. 

Transpose  p,  prt  =  a—p. 

Divide  by  pt,  r  =  *-^^.  (Formula  5.) 

P* 

Exercise  57. 
Solve  by  the  preceding  formulas : 

1.  The  sum  of  two  numbers  is  40,  and  their  difference  is 
10.     Find  the  numbers. 

2.  The  sum  of  two  angles  is  100°,  and  their  difference  is 
21°  30'.     Find  the  angles. 

3.  The  sum  of  two  angles  is  116°  24'  80",  and  their 
difference  is  56°  21'  44".     Find  the  angles. 


162  SCHOOL   ALGEBRA. 

4.  A  can  do  a  piece  of  work  in  6  days,  and  B  in  5  days. 
How  long  will  it  take  both  together  to  do  it  ? 

5.  Find  the  interest  of  $2750  for  3  years  at  4 J  per 
cent. 

6.  Find  the  interest  of  $950  for  2  years  6  months  at  5 
per  cent.    . 

7.  Find  the  amount  of  |2000  for  7  years  4  months 
at  6  per  cent. 

8.  Find  the  rate  if  the  interest  on  $680  for  7  months  is 
$35.70. 

9.  Find  the  rate  if  the  amount  of  $750  for  4  years  is 


10.  Find  the  rate  if  a  sum  of  money  doubles  in  16  years 
and  8  months. 

11.  Find  the  time  required  for  the  interest  on  $2130  to 
be  $436.65  at  6  per  cent. 

12.  Find  the  time  required  for  the  interest  on  a  sum  of 
money  to  be  equal  to  the  principal  at  5  per  cent. 

13.  Find  the  principal  that  will  produce  $  161 .25  interest 
in  3  years  9  months  at  8  per  cent. 

14.  Find  the  principal  that  will  amount  to  $1500  in  3 
years  4  months  at  6  per  cent. 

15.  How  much  money  is  required  to  yield  $  2000  interest 
annually  if  the  money  is  invested  at  5  per  cent? 

16.  Find  the  time  in  which  $640  will  amount  to  $1000 
at  6  per  cent. 

17.  Find  the  principal  that  will  produce  $100  per  month, 
at  6  per  cent. 

18.  Find  the  rate  if  the  interest  on  $700  for  10  months 
is  $25. 


CHAPTER  XI. 

SIMULTANEOUS    EQUATIONS    OF   THE    FIRST 
DEGREE. 

180.  If  we  have  two  unknown  numbers  and  but  one  rela- 
tion between  them,  we  can  find  an  unlimited  number  of 
pairs  of  values  for  which  the  given  relation  will  hold  true. 
Thus,  if  X  and  y  are  unknown,  and  we  have  given  only  the 
one  relation  a: -f- 3/ =  10,  we  can  assume  any  value  for  x, 
and  then  from  the  relation  x-\-y  =  V^  find  the  correspond- 
ing value  of  y.  For  from  a;  +  y  =  10  we  find  y  =  10  —  :r. 
If  X  stands  for  1,  y  stands  for  9  ;  if  a;  stands  for  2,  y  stands 
for  8  ;  if  a;  stands  for  —  2,  y  stands  for  12 ;  and  so  on  with- 
out end. 

181.  We  may,  however,  have  two  equations  that  express 
diffei^ent  relations  between  the  two  unknowns.  Such  equa- 
tions are  called  independent  equations.  Thus,  a;  +  y=10 
and  a;— y  =  2  are  independent  equations,  for  they  evidently 
express  different  relations  between  a;  and  y. 

182.  Independent  equations  involving  the  same  unknowns 
are  called  simultaneous  equations. 

If  we  have  two  unknowns,  and  have  given  two  independ- 
ent equations  involving  them,  there  is  but  one  pair  of  values 
which  will  hold  true  for  both  equations.  Thus,  if  in  §  181, 
besides  the  relation  a:  +  y  =  10,  we  have  also  the  relation 
a;  —  y  =  2,  the  only  pair  of  values  for  which  both  equations 
will  hold  true  is  the  pair  a:  =  6,  y  =  4. 

Observe  that  in  this  problem  x  stands  for  the  same  num- 
ber in  both  equations ;  so  also  does  y. 


164  SCHOOL   ALGEBRA. 

183.  Simultaneous  equations  are  solved  by  combining 
the  equations  so  as  to  obtain  a  single  equation  with  one 
unknown  number ;  this  process  is  called  elimination. 

There  are  three  methods  of  elimination  in  general  use : 

I.    By  Addition  or  Subtraction. 
II.    By  Substitution. 
III.    By  Comparison.       ^ 

184.  Elimination  by  Addition  or  Subtraction. 

(1)  -Solve:  5x-^y--^20}  (!) 

2;r  +  5^-39i  (2) 

Multiply  (1)  by  5,  and  (2)  by  3, 

25rr-153/  =  100  (3) 

6a; +  151/ =  117  (4) 

Add  (3)  and  (4),  31^  =217 

.-.  aj  =  7. 
Substitute  the  value  of  x  in  (2), 

14 +  52/ =  39. 
.-.  2/  =  5. 

In  this  solution  y  is  eliminated  by  addition. 


(2)   Solve:                6a;+35y-177) 
8a;-21y=    33  J 

(1) 

(2) 

Multiply  (1)  by  4,  and  (2)  by  3, 

24x  + 140^-708 

(3) 

24a;-   63y=    99 

(4) 

Subtract,                                   203  y= 609 

.••3/=3. 

Substitute  the  value  of  y  in  (2), 

8a; -63  =  33. 

.-.  a;  =12. 

In  this- solution  x  is  eliminated  by  subtraction. 


SIMULTANEOUS   EQUATIONS.  165 

186.   To  eliminate  by  addition  or  subtraction,  therefore, 

Multiply  the  equations  hy  such  numbers  as  will  make  the 
coefficients  of  one  of  the  unknown  numbers  equal  in  the 
resulting  equations. 

Add  the  resulting  equations,  or  subtract  one  from  the  other, 
according  as  these  equal  coefficients  have  unlike  or  like  signs. 

Note.  It  is  generally  best  to  select  the  letter  to  be  eliminated 
which  requires  the  smallest  multipliers  to  make  its  coefficients  equal ; 
and  the  smallest  multiplier  for  each  equation  is  found  by  dividing 
the  L.  C.  M.  of  the  coefficients  of  this  letter  by  the  given  coefficient  in 
that  equation.  Thus,  in  example  (2),  the  L.  G,  M.  of  6  and  8  (the  co- 
efficients of  x)  is  24,  and  hence  the  smallest  multipliers  of  the  two 
equations  are  4  and  3  respectively. 

Sometimes  the  solution  is  simplified  by  first  adding  the 
given  equations,  or  by  subtracting  one  from  the  other. 

(3)  a;  +  49y=    51  (1) 

i9x+      y=    99  (2) 

Add  (1)  and  (2),  50a;  +  50y  =  150  (3) 

Divide  (3)  by  50,  a;  +  3/  =  3.  (4) 

Subtract  (4)  from  (1),  48  y  =  48. 

•••  y  =  i. 

Subtract  (4)  from  (2),  48  a;  =  96. 

.-.  a;  =  2. 

Exercise  58. 

Solve  by  addition  or  subtraction : 

1.  5a;  +  2y  =  39|        4.  4a;  — 6y  =  26) 
2x-    y=    33  3a;-6y  =  15j 

•  2.   a:  +  3y  =  22")        5.   a:  +  2y  =  35') 
2:r-4y=  4j  Zx-2y^\l] 

8.  7a;-23/  =  ll)        6.   ^  +  4y  =  35') 
a;  -j-  5y  =  28  3  2a;  -  3y  =  26  3 


166  SCHOOL   ALGEBRA. 


3a;  +  5y  =  60|  11.      x+    2y=    9) 

X-  1y=    8)  ^x-    3y  =  903 

5a74-2y  =  36|  12.    ^x~    Sy^SO) 

2a7  +  3y  =  43j  3;r-   4y  =  173 

3^+7y=-50|  13.    7:^;-    2y=69j 

=  15  3  a;-10v=:39  3 


5a;-2y=15  3  a;-10y=:39 

2x-{-    y=    3|  14.    3a:+    7y=16 

7^  +  5y  =  2l3  2x-\-    5y  =  13 


186.   Elimination  by  Substitution. 

(1)   Solve:  5:^ +  4?/ =  32) 

4:r  +  3y  =  25) 

5a;  +  4y  =  32.  (1) 

4a;  +  3y-25.  (2) 

Transpose  4 y  in  (1),  5a;  =  32 -41/.  (3) 

Divide  by  coefficient  of  «,  a;  =  — ^^^^^ — ^-  (4) 

5 
Substitute  the  value  of  x  in  (2), 

■32 


(^^) 


32/ =  25. 


128 


i^  +  32/  =  25. 


5 

128-16y  +  153/  =  125, 
-y  =  -3. 
.'.  y  =  3. 
Substitute  the  value  of  y  in  (2), 

4a; +  9  =  25. 
.'.  a;  =  4. 

To  eliminate  by  substitution,  therefore, 

From  one  of  the  equations  obtain  the  value  of  one  of  the 
unknown  numbers  in  terms  of  the  other. 

Substitute  for  this  unknown  number  its  value  in  the  other 
equation,  and  reduce  the  resulting  equation. 


SIMULTANEOUS    EQUATIONS.  167 

Exercise  59. 

Solve  by  substitution : 

1.  2x-    7y=    0|  8.    Zx-    2y  =  28] 
Zx-    by  =  l\)  2x+    5y=63j 

2.  ^x-    5y=    4|  9.    2x~    3y  =  23| 
Zx-    2y  =  10j  5a;+    2y  =  29  3 

3.  2x-    3y-=    1|  10.    6a:-    7y  =  11  j 
2>x-    2y  =  29j  bx-    6y=    83 

4.  a;+      y  =  19|  11.    lx-\-    6y  =  20| 
2x+    7y  =  883  ■'2.X+    5y  =  323 

5.  2x-      y=    5|  '  12.      a;+    5y  =  37| 

:r4-    2y  =  25  3  3a:+    2y  =  46  3 

6.  19;r-15y  =  23|  13.    2>x-    7y  =  40| 


13a:-    5y  =  2l3  ^x-    3y=    9 

a;-f-10y  =  73|  14.    5a;+    9y  =  — : 

7a;-    2y=    73  3a;+lly  = 


187.   Elimination  by  Comparison. 
Solve:  2a;— 5y  =  66 


3a;  +  2y  =  233 

2a; -52/ =  66. 

(1) 

3a; +  23/ =  23. 

(2) 

Transpose  5y  in 

(1),  and  2y  in  (2), 

2a;  =  66  +  5y, 

(3) 

3a;  =  23-2y. 

(4) 

Divide  (3)  by  2, 

2 

(5) 

Divide  (4)  by  3, 

:,._23-2^. 
3 

(6) 

Equate  the  values  of  x. 

66  +  5v     23 -2v 
2                3 

(V) 

168 

SCHOOL   ALGEBRA. 

Reduce 

(7). 

198  +  15y  = 
19  J/ = 

•*•  y  = 

46 -4y, 
■  -  152. 
=  -8. 

Substitute  the 

value 

of  3/  in  (1), 

2a:  +  40  = 

=  66. 

.'.    X  = 

=  13. 

188.   To  eliminate  by  comparison,  therefore, 

From  each  equation  obtain  the  value  of  one  of  the  unknowv 
numbers  in  terms  of  the  other. 

Form  an  equation  from  these  equal  values  and  reduce  the 
equation. 

Exercise  60. 
Solve  by  comparison : 

1.  x+      y  =  30|  9.      2x-    Zy=      1) 

Zx-   23/  =  25j  5a;+    2y  =  126j 

2.  7:r+    3y=70|  lO.    mx-    9y=      1) 

bx-    4y=    73  Ix-      y=      3j 

3.  9a;+    4y  =  54")  11.        x-\-'2Xy^      2| 
4^+    92/  =  89j  27y+    2;r=    19  j 

4.  7:r+    2y  =  63")  12.    10a;+    3y  =  174 
8a;-      y==    3j  3:r  +  10y  =  125 

5.  2a:-33y=29|  13.     6a;-13y=      2) 
3a;--47y-:46  j  5a7-12y=      4) 

6.  2a;-      y=    9|  14.      2a;  +      y  =  108 ) 
5a;-    3y  =  143  10a;+   2y=    60  3 

7.  11a;-    7y-    6j  15.      3a;-    5y=      5| 

9a;-   5y  =  10j  7a;4-      y  =  2653 

8.  5a;  +  9y  =  188)  16.    12a;+    7y  =  176) 
13a; -2y=    57  3  3y-19a;=      33 


SIMULTANEOUS    EQUATIONS. 


169 


189.    Each   equation   must   be   simplified,  if  necessary, 
before  the  elimination. 


Solve:  f^-i(y+l)  =  l) 

i(a:+l)  +  |(y-l)  =  9j 

|x-i(y+l)=l. 

H^  +  i)  +  l(y-i)  =  9. 

Multiply  (1)  by  4,  and  (2)  by  12, 

3a;-22/-2  =  4, 
4a; +  4 +  92/ -9  =  108. 
From  (3),  3a;-2y  =  6. 

From  (4),  4a;  +  9y  =  113. 

Multiply  (5)  by  4,  and  (6)  by  3, 

12a:-    8y=    24 
12a; +  272/ =  339 

352/  =  315 
.-.  2/  =  9 


(1) 

(2) 

(3) 
(4) 
(5) 
(6) 


Substitute  value  of  y  in  (1), 


8. 


Solve : 

1.    ^  +  ^  =  ^] 
3      2      3  I 


Exercise  61. 


^4-^  =  2 
2^3      6 


2.  ^^4-tl^-^9 

3  2 

2^     9 

3.   ±—JL  =  x  —  y 

40  ^ 


2a;-y 


3       ^^^--2 


4.    ^Zll-^^^  =  2 
^      3 


3a;-5y      3^2:r  +  y 
2       ^  5 

3_a;-2y^g^ 
4  2^3 


6. 


x  —  y 
7a;- 13 


3y-5 


=  4 


170 


SCHOOL   ALGEBRA. 


3  4  5 

4  3     ~      2 


8.    ^-±M±1  =  2] 
2x-y+l 

x  —  y  +  S 


9     ^-2      10-:r_y-10 
5  3  4 

2?i+i  ^  4a:  +  y  +  26 
3"  8 


10. 


2x-y  +  S  _x  —  2y-\-19 
3  4 

3y-4y  +  3  ^  2y-4ar  +  21 
■       4  3 


T_-t^  _  2^  — y , 


11.    ^^-_r.r_^^3y-5 

5^-_7     4^_-3_^g_5^ 
2  6 


12.    ! ^  =  4y  — 3a; 

13  ^ 

..p  ,  5a;-6y^3(a:  +  2y) 
"^6  '4 


13  ^±g^2-^y~^ 

2  6 

2x  +  y     9a;— 7^  7y  — 4a; +  36 
2  8  16 


14. 


a;  +  2y  +  3   _5y-4a;-6 
13  3 

6a;— 5y  +  4_3a?  +  2y+l 
3  19 


SIMULTANEOUS    EQUATIONS. 


171 


15. 


a;  +  y  _  5 
y-x      3 

7 


16. 


y  —  4  __  a:  +  1 


4  10 

^  ,  y  +  2 

6"^     5 


3i 


17.   ^^+2.=.3(y-l) 
5a;  +  6y     4a7  — 3y_     _ 


10 


y 


18. 


4a;-3y-7_^9a;-4y-25 
5  30 

y-1  ■  10a:-3y-20_3a7+2y+3 
3     "^  20  30 

Note.   In  solving  the  following  problems  proceed  as  in  \  170 
19.    6.y  +  5      4rr-5y  +  3_9y-4 


8  4.r  — 2y 

•£j-_3  ,  a;  —  3y _6a;  —  1 
4      "^   7-a;  ~      3 


12 


20.    X 


2^::^  =  ^-9i' 


23 -a; 


y\- 


-    3 


a;- 18 


y-^i 


21  4a7+7  .  5a;  — 4y_17  +  8a? 

3  "^  2a;  +  8  6 

5a?-12     4a;-6y-13^10a:-53 

4  2a;  — 3y  8 

22  7  +  8a;      3(a;  -  2y)  _  11  +  4a;      ^ 

10  2(a;-4)  5 

3(2y  +  3)_6y  +  21       3y  +  5a; 
4  4  2(2y-3)J 


172  SCHOOL    ALGEBRA. 

190.   Literal  Simultaneous  Equations. 

Solve  :  ax  -\-  b^/  ==  c  "l 

a^x  -\-  Vy  =  c' ) 

Note.  The  letters  a^,  y  are  read  a  prime,  h  prime.  In  like  man- 
ner, a^^,  a^^^  are  read  a  second,  a  third,  and  ai,  Oj,  as,  are  read  a  sub 
one,  a  sub  two,  a  sub  three.  It  is  sometimes  convenient  to  represent 
different  numbers  that  have  a  common  property  by  the  same  letter 
marked  by  accents  or  suffixes.  Here  a  and  a^  have  a  common 
property  as  coefficients  of  x. 

ax   +by   ==  c.  (1) 

a^x  +  ¥y  =  c\  (2) 

To  find  the  value  of  y,  multiply  (1)  by  a'',  and  (2)  by  a, 

aa^x  i-  a^by  =  a^c 

aa^x  -4-  ab^v  =  ac^ 


a^by  —  ab^y  =  a^c  —  ac^ 
a^c  —  ac^ 


^     a'b-ay 

To  find  the  value  of  x,  multiply  (i)  by  b\  and  (2)  by  b, 
aVx  +  bb^y  =  Vc 
a^bx  +  bVy  =  be^ 

aVx  —  a^bx  =  Vc  —  b(/ 

^^b'c-bc^ 

"^  a¥-a'b 

Exercise  62. 


Solve : 
+ 

y 


1*   x-\-y  =  8  \  6.    hx-\-ay=^ahc\ 

x  —  y  —  dy  x  =  dy  ) 

2.    mx  -\-ny  —r  )  ^.   hx  ■{- ay  —1\ 
mx  +  n'y  =  r'  )  h^x  —  a'y  =  1  3 

Z,    ax  -hy  ^^c')  7.    ^hx-\-2ay  =  '6ab\ 
a'x-\-b'y  =  c*)  4:bx-Say  =  lab\ 

4.     x~      y  =  mn)  8.    2x  —  Sy  =  a~b') 
ex ~\- aby  =  ms  )  2>x —  2y  —  a-\'h) 


SIMULTANEOUS   EQUATIONS. 


173 


bx 


cy 


bx  -{-  07/ =  a -\- b 


10 

x  +  m_a 

y-n      b 

^ 

bx  ■\-  ay  ~  c 

11. 

a       b 

2x_y__\ 
a       6      3 

12. 

^    1     y 

6=2 

a-\-b     a  — 

x-\-y  =  2a 

13. 

2x     ^_Z 
ZaTZb       I 

14.  a:r  +  Jy  =  c  *) 
bx  -{-  ay  ^=  c  ) 

15.  3a'  +    aa7  =  5'  +  6y) 

ax-{-2by  =  d  ) 


16. 


2/    _     1 


17. 


x—y=a— b 


a-\-b      a—b      a-\-b 
a-f  6      a  — 6      a  —  b 

^  =  1 
1 


m—a ' m—b 


w  — a      71 


-^> 


191.  Fractional  simultaneous  equations,  of  which  the  de- 
nominators are  simple  expressions  and  contain  the  unknown 
numbers,  may  be  solved  as  follows : 


(1)    Solve; 


"We  have 


and 


To  find  the  value  ofy. 
Multiply  (1)  by  c, 


a  ,  b 

— \--  =  ni 

X     y 

c  ,  d 

X     y 

a  ,   h 
X     y 

c     d 
-  +  -  -  n. 
X     y 


ac     he 

— h  —  =  cm. 

X      y 


(1) 
(2) 

(3) 


174  SCHOOL    ALGEBRA. 

Multiply  (2)  by  a,  ^^^l^an.  .(4) 

X       y 

Subtract  (4)  from  (3),  ^^"^^  =cm-  an. 

y 

Multiply  both  sides  by  y,    hc  —  ad='  {cm  —  an)y. 

be  —  ad 

'■  y- 


cm  — an 


X. 


(5) 


To  find  the  value  of 

Multiply  (1)  by  ^.  ^  +  ^=^m. 

X       y 

Multiply  (2)  by  6.  h  +  M^hn.  (6) 

X      y 

Subtract  (6)  from  (5),  ad-bc^  ^^  _  ^^^ 

X 

Multiply  both  sides  by  x,   ad—bG  =  {dm  —  bn)x. 

ad  —  be 

. .  X  = —  • 

dm  —  bn 

(2)   Solve:  #-  +  #-  =  7 

6x      lOy 

We  have  A  +  A=.7.  n) 

Sx     5y  ^  ^ 

and  -± —  =  3.  ^2) 

6x     lOy  "^  ' 

Multiply  (1)  by  15,  the  L.C.M.  of  3  and  5,  and  (2)  by  30, 

?5  +  5  =  105.  (3) 

X      y 

---  =  90.  (4) 

X      y 

Multiply  (4)  by  2,  and  add  the  result  to  (3), 

55-286. 

X 

'   x=^l 
,.x^^ 

Substitute  the  value  of  a?  in  (1),  and  we  get 


SIMULTANEOUS   EQUATIONS. 


176 


oh 

re: 

Exercise  63. 

1. 

X     y 

7. 

1,1           1 

X      y 

, 

X     y 

2. 

5  +  l!  =  49^ 
X      y 

8. 

^+l=i' 

^    y 

^    y 

^    y 

3. 

5+5=3     1 
X     y 

«            9. 

m  ,  n 
X      y 

^    y       J 

^+i  =  J 

X     y 

4. 

2  +  5=19] 

^    y 

10. 

a      b      ac^ 
x~y~  b 

b      a  __bc 
X     y      a  ^ 

5. 

L+!r'ii 

11. 

ax      by 

^ 

5        4       11 
4^x     by      20 

. 

ax     by 

> 

6. 

2i^3y 

12. 

ox     ay 

Ax      by 

9 

-b' 

176  SCHOOL   ALGEBHA. 

192,  If  three  simultaneous  equations  are  given,  involv- 
ing three  unknown  numbers,  one  of  the  unknowns  must  be 
eliminated  between  two  pairs  of  the  equations;  then  a 
second  unknown  between  the  two  resulting  equations. 

Likewise,  if  four  or  more  equations  are  given,  involving 
four  or  more  unknown  numbers,  one  of  the  unknowns  must 
be  eliminated  between  three  or  more  pairs  of  the  equations ; 
then  a  second  between  the  pairs  that  can  be  formed  of  the 
resulting  equations ;  and  so  on. 

Note.  The  pairs  chosen  to  eliminate  from  must  be  independent 
pairs,  so  that  each  of  the  given  equations  shall  be  used  in  the  process 
of  the  eliminations. 

Solve:  2a;  — 3y  +  42;=    A^-\  (1) 

3a:  +  5y-70  =  12[  (2) 

Eliminate  z  between  the  equations  (1)  and  (3). 

Multiply  (1)  by  2,  4a;  -  Gy  +  82  =    8  (4) 

(3)  is  5  a;  —    y  —  8  z  =   5 

Add,  2x-1y  ^Ts  (5) 

Eliminate  2  between  the  equations  (1)  and  (2). 

Multiply  (1)  by  7,       14  a;- 21^/ +  28z  =  28 

Multiply  (2)  by  4,       12a;  +  20^/ -  28z  =  48 

Add,  26  a;-      y  ^Tq  (6) 

We  now  have  two  equations  (5)  and  (6)  involving  two  unknowns, 
X  and  y. 

Multiply  (6)  by  7,  182  a;  -  7  3/  =  532  (7) 

(5)  is  9a; -72/=    13 

Subtract  (5)  from  (7),  173  a;  =519 

.-.  a;  =  3. 
Substitute  the  value  of  x  in  (6),  78  —  3/  =  76. 

.-.2/ =  2. 
Substitute  the  values  of  x  and  y  in  (1), 

6_6  +  4z  =  4. 

.-.  2  =  1. 


SIMULTANEOUS    EQUATIONS.  177 


Exercise  64. 


1.  a;-fy-  8  =  0]  10.  5a;+ 2y  —  202  =  20 
y-fz-28  =  0>  3a;— 63/+  72  =  51 
a;-fz-14  =  oJ  4a;  +  8y—    9z  =  53 

2.  4^  +  3y  +  22  =  251  11.      a;-^2y+10z=   44 
3:r-2y  +  5z  =  20  >  3a;  +  3y+    7z  =  384 

10a:-5y  +  3z  =  17  J  2x-\-    y+      z  =  256 

3.  5a;-2y-2z=12]  12.    10a;=    y  +  4z  +  56 

x^    2/+    z=    8V  3y  =  2a;+32  — 98 

7a;+3y  +  42  =  42  J  2z  =    a:— 3y— 18 

4.  a;—    y+    2  =  11]  13.      3a7  —  5y  — 22  =  14 
3a:  +  3y-22  =  60  y  5:i;_8y-    2=12 

lOa;  — 5y  — 3z=    oj  a:  — 3y-32=    1 

5.  lOa;—    y  +  32  =  42]  14.      2a:  +  3y+    2  =  31 

7a;  +  2y+    2  =  51  >  x-    y+32=13 

3a7  +  3y-    2  =  24j  10y  +  5aj-2z  =  48 

6.  5a7  +  2y  — 32  =  160  j  15.  2x-\-  3y-  42  =  1 
3a7+9y  +  82  =  115  ^  10a;-  6y+122--=6 
2a;~3y  — 5z=    45)  a;  +  12y+    22  =  5 

7.  6a;  — 2y+52  =  53  1  16.  3a;+  6y  +  22  =  3 
5a;  +  3y+72  =  33  [  12y+42      ^  —  " 

x-\-    y+    z=    b)  9a;+18y 


+  22  =  3] 
-6a;=2[ 
-42  =  4J 


8.  3a;-3y  +  4z  =  20|  17.  2a;+  y  +  22  =  3] 
6a;+2y— 72=  5^  5y  — 4a;  — 42  =  l[ 
2a;-    y  +  82  =  45)  3.r  +  9y+    z  =  9J 

9.  2a;+7y+ 102  =  25)  18.    3a;  +  2y+      z  =  20^ 

1 

2" 


a;+    y—      2=    9^  2a;-    y4-    32  =  26 

7a;-7y-ll2  =  73]  x-\-    y+10z  =  55 


178 


SCHOOL    ALGEBRA. 


19.    1-?+    4  =  0 
X     y 


+    1  =  0 


1 

y    2 


2  +  §_14  =  0 

Z        X 


23.    1  +  1-1  =  0 

X     y      a 

1+1-^0 
X     z      h 

1+1-1=0 
y     z      c 


20.   i+i  +  i     =36 

X     y     z 

1  +  3_1     =28 
X     y     z 


24.   1+1+1  =  62 
|  +  f+|  =  47 


2X.   l  +  ?-5=   1 

5+^+5  =  24 
a;  '  y  '  z 

!-?  +  H=i4 
37     y     z 


25.    ?  +  l-?  =  0 


-2  =  0 


3_2 

z      y 


22.    i-?  =  -l 

a;     y      20 

z  a;  15 
4_5_J^ 
2     y.    12^ 


15     4  ,  5 

26. h-=    38 

a;      y     z 

?  +  5  +  l?=    61 
X     y       z 

§-5  +  ^=161 

a;     y       z 


CHAPTER   XII. 

PROBLEMS  INVOLVING  TWO  OR  MORE  UNKNOWN 
NUMBERS. 

193.  It  is  often  necessary  in  the  solution  of  problems  to 
employ  two  or  more  letters  to  represent  the  numbers  to  be 
found.  In  all  cases  the  conditions  must  be  sufficient  to 
give  just  as  many  equations  as  there  are  unknown  numbers 
employed. 

194.  If  there  are  more  equations  than  unknown  numbers, 
some  of  them  are  superfluous  or  inconsistent ;  if  there  are 
fewer  equations  than  unknown  numbers,  the  problem  is 
indeterminate. 

(1)  If  A  gives  B  $10,  B  will  have  three  times  as  much 
money  as  A.  If  B  gives  A  f  10,  A  will  have  twice  as  much 
money  as  B.     How  much  has  each  ? 

Let  X  =  number  of  dollars  A  has, 

and  y  =  number  of  dollars  B  has. 

Then,  after  A  gives  B  $  10, 

a;  —  10  =  the  number  of  dollars  A  has, , 
y  +  10  ==  the  number  of  dollars  B  has. 
.-.  y  +  10  =  3(a;-10).  (1) 

If  B  gives  A  f  10, 

a;  +  10  =  the  number  of  dollars  A  has, 
y  —  10  =  the  number  of  dollars  B  has. 
.-.  a;  +  10  =  2(y-10).  (2) 

From  the  solution  of  equations  (1)  and  (2),  x  =•  22,  and  y  =  2Q. 
Therefore  A  has  $22,  and  B  has  $26. 


180  SCHOOL    ALGEBRA. 

(2)  If  the  smaller  of  two  numbers  is  divided  by  the 
greater,  the  quotient  is  0.21,  and  the  remainder  0.0057; 
but  if  the  greater  is  divided  by  the  smaller,  the  quotient 
is  4  and  the  remainder  0.742.     Find  the  numbers. 


(1) 

(2) 

(3) 
(4) 
(5) 


Let 
and 

X  =  the  greater  number, 
y  =  the  smaller  number. 

Then 

y- 0.0057  _ 

X 

and 

a?- 0.742     ^ 

y 

.'.  y  -  0.21  X  =  0.0057, 

a: -4-2/ =  0.742. 

Multiply  (3)  by  4,         4  y  -  0.84  x  =  0.0228 
(4)  is                          -42/+         a^  =  0.742 
By  adding,                              0.16a;  =  0.7648 

.-.  a;  =4.78. 
Substituting  the  value  of  x  in  (4), 

-4y  =  - 4.038, 
::  y  =  1.0095. 

Exercise  65. 

1.  If  A  gives  B  1 100,  A  will  then  have  half  as  much 
money  as  B ;  but  if  B  gives  A, $100,  B  will  have  one-third 
as  much  as  A.     How  much  has  each  ? 

2.  If  the  greater  of  two  numbers  is  divided  by  the 
smaller,  the  quotient  is  4  and  the  remainder  0.37;  "but  if 
the  smaller  is  divided  by  the  greater,  the  quotient  is  0.23 
and  the  remainder  0.0149.     Find  the  numbers. 

3.  A  certain  number  of  persons  paid  a  bill.  If  there 
had  been  10  persons  more,  each  would  have  paid  $2  less; 
but  if  there  had  been  5  persons  less,  each  would  have  paid 
$2.50  more.  Find  the  number  of  persons  and  the  amount 
of  the  bill. 


FBOBLEMS. 


181 


4.  A  train  proceeded  a  certain  distance  at  a  uniform 
rate.  If  the  speed  had  been  6  miles  an  hour  more,  the  time 
occupied  would  have  been  5  hours  less ;  but  if  the  speed  had 
been  6  miles  an  hour  less,  the  time  occupied  would  have 
been  7|-  hours  more.     Find  the  distance. 

Hint.  If  x  =  the  number  of  hours  the  tram  travels,  and  y  the 
number  of  miles  per  hour,  then  xy  =•  the  distance. 

5.  A  man  bought  10  cows  and  50  sheep  for  $750.  He 
sold  the  cows  at  a  profit  of  10  per  cent,  and  the  sheep  at  a 
profit  of  30  per  cent,  and  received  in  all  $875.  Find  the 
average  cost  of  a  cow  and  of  a  sheep. 

6.  It  is  40  miles  from  Dover  to  Portland.  A  sets  out 
from  Dover,  and  B  fron^^ Portland,  at  7  o'clock  a.m.,  to  meet 
each  other.  A  walks  at  the  rate  of  3^  miles  an  hour,  but 
stops  1  hour  on  the  way ;  B  walks  at  the  rate  of  2^  miles  an 
hour.  At  what  time  of  day  and  how  far  from  Portland  will 
they  meet? 

7.  The  sum  of  two  numbers  is  35,  and  their  difference 
exceeds  one-fifth  of  the  smaller  number  by  2.  Find  the 
numbers. 

8.  If  the  greater  of  two  numbers  is  divided  by  the 
smaller,  the  quotient  is  7  and'the  remainder  4 ;  but  if  three 
times  the  greater  number  is  divided  by  twice  the  smaller, 
the  quotient  is  11  and  the  remainder  4.    Find  the  numbers. 

9.  If  3  yards  of  velvet  and  12  yards  of  silk  cost  $60, 
and  4  yards  of  velvet  and  5  yards  of  silk  cost  $58,  what  is 
the  price  of  a  yard  of  velvet  and  of  a  yard  of  silk  ? 

10.  If  5  bushels  of  wheat,.  4  of  rye,  and  3  of  oats  are  sold 
for  $  9 ;  3  bushels  of  wheat,  5  of  rye,  and  6  of  oats  for  $  8. 75 ; 
and  2  bushels  of  wheat,  3  of  rye,  and  9  of  oats  for  $  7.25  ; 
what  is  the  price  per  bushel  of  each  kind  of  grain  ? 


182  SCHOOL   ALGEBRA. 

Note  I.  A  fraction  the  terms  of  which  are  unknown  may  be  repre- 
sented by  -. 

y 

Ex.  A  certain  fraction  becomes  equal  to  I-  if  2  is  added 
to  its  numerator,  and  equal  to  I-  if  3  is  added  to  its  denomi- 
nator.    Find  the  fraction. 

Let  -  =  the  required  fraction. 

y 

Then  ^^  =  J. 

y 

and  ^ 


y  +  3    ' 

The  solution  of  these  equations  gives  7  for  x,  and  18  for  y. 
Therefore  the  required  fraction  is  •^. 

11.  A  certain  fraction  becomes  equal  to  -J  if  3  is  added 
to  its  numerator  and  1  to  its  denominator,  and  equal  to  \ 
if  3  is  subtracted  from  its  numerator  and  from  its  denomi- 
nator.    Find  the  fraction. 

12.  A  certain  fraction  becomes  equal  to  -j^  if  1  is  added 
to  double  its  numerator,  and  equal  to  -^  if  3  is  subtracted 
from  its  numerator  and  from  its  denominator.  Find  the 
fraction. 

13.  Find  two  fractions  with  numerators  11  and  5  respec- 
tively, such  that  their  sum  is  l-f,  and  if  their  denominators 
are  interchanged  their  sum  is  2-J. 

14.  There  are  two  fractions  with  denominators  20  and 
16  respectively.  The  fraction  formed  by  taking  for  a  nu- 
merator the  sum  of  the  numerators,  and  for  a  denominator 
the  sum  of  the  denominators,  of  the  given  fractions,  is  equal 
to  J;  and  the  fraction  formed  by  taking  for  a  numerator 
the  difference  of  the  numerators,  and  for  a  denominator  the 
difference  of  the  denominators  of  the  given  fractions,  is  equal 
to  \.     Find  the  fractions. 


PROBLEMS.  183 

Note  II.  A  number  consisting  of  two  digits  which  are  unknown 
may  be  represented  by  lOx  +  y,  in  which  x  and  y  represent  the  digits 
of  the  number.  Likewise,  a  number  consisting  of  three  digits  which 
are  unknown  may  be  represented  by  100  a:  +  10 y  +  z,  in  which  x,  y, 
■and  2  represent  the  digits  of  the  number.  For  example,  the  expres- 
sion 364  means  300  +  60  +  4 ;  or,  100  times  3  +  10  times  ^^  +  4. 

Ex.  The  sum  of  th6  two  digits  of  a  number  is  10,  and  if 
18  is  added  to  the  number,  the  digits  will  be  reversed. 
Find  the  number. 


Let 

X  =  tens'  digit. 

and 

y  =>  units'  digit. 

Then 

10  a?  +  y  ==■  the  number; 

Hence 

a;  +  2/  =  10, 

(1) 

and 

10  a; 

+  y  +  18='10y  +x. 

(2) 

From  (2), 

9a;-9y  =  -18, 

or 

x-y  =^-  2. 

(3) 

Add  (1)  and  (3), 

2a;  =    8, 

and  therefore 

a;=    4. 

Subtract  (3)  from  (1), 

22/ =  12. 

and  therefore 

y-  6. 

Therefore  the  number  is  46.     . 

16.  The  sum  of  the  two  digits  of  a  number  is  9,  and  if  27 
is  subtracted  from  the  number,  the  digits  will  be  reversed. 
Find  the  number. 

16.  The  sum  of  the  two  digits  of  a  number  is  9,  and  if 
the  number  is  divided  by  the  sum  of  the  digits,  the  quotient 
is  5.     Find  the  number. 

17.  A  certain  number  is  expressed  by  two  digits.  The 
sum  of  the  digits  is  11.  If  the  digits  are  reversed,  the  new 
number  exceeds  the  given  number  by  27.    Find  the  number. 

18.  A  certain  number  is  expressed  by  three  digits.  The 
sum  of  the  digits  is  21.  The  sum  of  the  first  and  last  digits 
is  twice  the  middle  digit.  If  the  hundreds*  and  tens'  digits 
are  interchanged,  the  number  is  diminished  by  90.  Find 
the  number. 


184  SCHOOL    ALGEBRA. 

19.  A  certain  number  is  expressed  by  three  digits,  the 
units'  digit  being  zero.  If  the  hundreds'  and  tens'  digits  are 
interchanged,  the  number  is  diminished  by  180.  If  the 
hundreds'  digit  is  halved,  and  the  tens'  and  units'  digits  are 
interchanged,  the  number  is  diminished  by  336.  Find  the 
number. 

20.  A  number  is  expressed  by  three  digits.  If  the  digits 
are  reversed,  the  new  number  exceeds  the  given  number  by 
99.  If  the  number  is  divided  by  nine  times  the  sum  of  its 
digits,  the  quotient  is  3.  The  sum  of  the  hundreds'  and 
units'  digits  exceeds  the  tens'  digit  by  1.    Find  the  number. 

Note  III.   If  a  boat  moves  at  the  rate  of  x  miles  an  hour  in  still 
water,  and  if  it  is  on  a  stream  that  runs  at  the  rate  of  y  miles  an 
nour,  t  en      ^j  +  3/  represents  its  rate  down  the  stream, 
x  —  y  represents  its  rate  wp  the  stream. 

21.  A  boatman  rows  20  miles  down  a  river  and  back  in 
8  hours.  He  finds  that  he  can  row  5  miles  down  the  river 
in  the  same  time  that  he"  rows  3  miles  up  the  river.  Find 
the  time  he  was  rowing  down  and  up  respectively. 

22.  A  boat's  crew  which  can  pull  down  a  river  at  the 
rate  of  10  miles  an  hour  finds  that  it  takes  twice  as  long  to 
row  a  mile  up  the  river  as  to  row  a  mile  down.  Find  the 
rate  of  their  rowing  in  still  water  and  the  rate  of  the 
stream. 

23.  A  boatman  rows  down  a  stream,  which  runs  at  the 
rate  of  2^  miles  an  hour,  for  a  certain  distance  in  1  hour 
and  30  minutes ;  it  takes  him  4  hours  and  30  minutes  to 
return.  Find  the  distance  he  pulled  down  the  stream  and 
his  rate  of  rowing  in  still  water. 

Note  IV.  It  is  to  be  remembered  that  if  a  certain  work  can  be 
done  in  x  units  of  time  (days,  hours,  etc.),  the  part  of  the  work  done 
in  out  unit  of  time  will  be  represented  by  — 


PROBLEMS.  185 

Ex.  A  cistern  has  three  pipes,  A,  B,  and  0.  A  and  B 
will  fill  the  cistern  in  1  hour  and  10  minutes,  A  and  0  in 
1  hour  and  24  minutes,  B  and  C  in  2  hours  and  20  minutes. 
How  long  will  it  take  each  pipe  alone  to  fill  it  ? 

1  hour   and  10  minutes  =    70  minutes. 

1  hour  and  24  minutes  =    84  minutes. 

2  hours  and  20  minutes  =  140  minutes. 
Let                X  =  number  of  minutes  it  takes  A  to  fill  it, 

y  =  number  of  minutes  it  takes  B  to  fill  it, 
and  z  =  number  of  minutes  it  takes  C  to  fill  it. 

Then    _,_,_  =  the  parts  A,  B,  and  C  can  fill  in  one  minute 
^  respectively, 

and  _  4-  _  =  the  part  A  and  B  together  can  fill  in  one  minute. 

«    y 

But  —  «-  the  part  A  and  B  together  can  fill  in  one  minute. 

a;     2/     70  ^  ' 

In  like  manner,  -  +  -  =  — ,  (2) 

ar     z      84  ^  ' 

and  -  +  -  =  lk  <3) 

V     z     140 


Add,  and  divide  by  2,        1  +  1+1  =  JL  (4) 

X     y     z      60 


Subtract  (1)  from  (4), 
Subtract  (2)  from  (4), 
Subtract  (3)  from  (4), 


1  =  J_. 

z      420* 

1       1 
y^210 
1_  J_ 

X     105' 
Therefore  x,y,z  =  105,  210,  420,  respectively. 

Hence  A  can  fill  it  in  1  hour  and  45  minutes,  B  in  3 
hours  and  30  minutes,  and  0  in  7  hours. 

24.  A  and  B  can  do  a  piece  of  work  together  in  3  days, 
A  and  C  in  4  days,  B  and  0  in  4J-  days.  How  long  will  it 
take  each  alone  to  do  the  work  ? 


186  SCHOOL.  ALGEBRA, 

26.  A  and  B  can  do  a  piece  of  work  in  2J  days,  A  and 
C  in  3-|-  days,  B  and  0  in  4  days.  How  long- will  it  take 
each  alone  to  do  the  work  ? 

26.  A  and  B  can  do  a  piece  of  work  in  a  days,  A  and  C 
in  b  days,  B  and  0  in  cdays.  How  long  will  it  take  each 
alone  to  do  the  work  ? 

Note  V.  If  x  represents  the  number  of  linear  units  in  the  lengthy, 
and  y  in  the  width,  of  a  rectangle,  xy  will  represent  the  number  of 
its  units  of  surface,  the  surface  unit  having  the  same  name  as  the 
linear  unit  of  its  side. 

27.  If  the  length  of  a  rectangular  field  were  increased 
by  5  yards  and  its  breadth  increased  by  10  yards,  its  area 
would  be  increased  by  450  square  yards  ;  but  if  its  length 
were  increased  by  5  yards  and  its  breadth  diminished  by 
10  yards,  its  area  would  be  diminished  by  350  square  yards. 
Find  its  dimensions. 

28.  If  the  floor  of  a  certain  hall  had  been  2  feet  longer 
and  4  feet  wider,  it  would  have  contained  528  square  feet 
more  ;  but  if  the  length  and  width  were  each  2  feet  less,  it 
would  contain  816  square  feet  less.     Find  its  dimensions. 

29.  If  the  length  of  a  rectangle  was  4  feet  less  and  the 
width  3  feet  more,  the  figure  would  be  a  square  of  the  same 
area  as  the  given  rectangle.  Find  the  dimensions  of  the 
rectangle. 

Note  VI.  In  considering  the  rate  of  increase  or  decrease  in  quan- 
tities, it  is  usual  to  take  100  as  a  common  standard  of  reference,  so 
that  the  increase  or  decrease  is  calculated  for  every  100,  and  there- 
fore called  per  cent. 

It  is  to  be  observed  that  the  representative  of  the  number  result- 
ing after  an  increase  has  taken  place  is  100  +  increase  per  cent;  and 
after  a  decrease,  100  —  decrease  per  cent. 

Interest  depends  upon  the  time  for  which  the  money  is  lent,  as 


PROBLEMS.  187 

well  as  upon  the  rate  per  cent  charged ;  the  rate  per  cent  charged 
being  the  rate  per  cent  on  the  principal  for  one  year.     Hence, 

Simple  interest  =  Principal  X  Rate  per  cent  X  Time^ 
^  100 

where  Time  means  number  of  years  or  fraction  of  a  year. 
Amount  =  Principal  +  Interest. 

In  questions  relating  to  stocks,  100  is  taken  as  the  representative 
of  the  stock,  the  price  represents  its  market  value,  and  the  per  cent 
represents  the  interest  which  the  stock  bears.  Thus,  if  six  per  cent 
stocks  are  quoted  at  108,  the  meaning  is,  that  the  price  of  1 100  of 
the  stock  is  $108,  and  that  the  interest  derived  from  $100  of  the 
stock  will  be  y^^  of  $  100,  that  is,  $  6  a  year.  The  rate  of  interest  on 
the  money  invested  will  be  }^§  of  6  per  cent. 

30.  A  man  has  $10,000  invested.  For  a  part  of  this 
sum  he  receives  5  per  cent  interest,  and  for  the  rest  6  per 
cent;  the  income  from  his  5  per  cent  investment  is  $60 
more  than  from  his  6  per  cent.  How  much  has  he  in  each 
investment  ? 

31.  A  sum  of  money,  at  simple  interest,  amounted  in  4 
years  to  $  29,000,  and  in  5  years  to  $30,000.  Find  the  sum 
and  the  rate  of  interest. 

32.  A  sum  of  money,  at  simple  interest,  amounted  in  10 
months  to  $2100,  and  in  18  months  to  $2180.  Find  the 
sum  and  the  rate  of  interest. 

33.  A  person  has  a  certain  capital  invested  at  a  certain 
rate  per  cent.  Another  person  has  $2000  more  capital, 
and  his  capital  invested  at  one  per  cent  better  than  the 
first,  arid  he  receives  an  income  of  $  150  greater.  A  third 
person  has  $3000  more  capital,  and  his  capital  invested  at 
two  per  cent  better  than  the  first,  and  he  receives  an  income 
of  $280  greater.  Find  the  capital  of  each  and  the  rate  at 
which  it  is  invested. 


188  SCHOOL    ALGEBRA. 

34.  A  sum  of  money,  at  simple  interest,  amounted  in  m 
years  to  c  dollars,  and  in  n  years  to  d  dollars.  Find  the 
sum  and  the  rate  of  interest. 

35.  A  sum  of  money,  at  simple  interest,  amounted  in  m 
months  to  a  dollars,  and  in  n  months  to  b  dollars.  Find 
the  sum  and  the  rate  of  interest. 

36.  A  person  has  $18,375  to  invest.  He  can  buy  3  per 
cent  bonds  at  75,  and  5  per  cent  bonds  at  120.  How  much 
of  his  money  must  he  invest  in  each  kind  of  bonds  in  order^ 
to  have  the  same  income  from  each  investment? 

Hint.  Notice  that  the  3  per  cent  bonds  at  75  pay  4  per  cent  on 
the  money  invested,  and  5  per  cent  bonds  at  120  pay  4|-  per  cent. 

37.  A  man  makes  an  investment  at  4  per  cent,  and  a 
second  investment  at  4|-  per  cent.  His  income  from  the 
two  investments  is  $715.  If  the  first  investment  had  been 
at  4|-  per  cent  and  the  second  at  4  per  cent,  his  income  would 
have  been  $  730.     Find  the  amount  of  each  investment. 

(1)  In  a  mile  race  A  gives  B  a  start  of  20  yards  and 
beats  him  by  30  seconds.  At  the  second  trial  A  gives  B  a 
start  of  32  seconds  and  beats  him  by  Q^^  yards.  Find  the 
number  of  yards  each  runs  a  second. 

Let  X  ==  number  of  yards  A  runs  a  second, 

and  ,  y  =  number  of  yards  B  runs  a  second. 

Since  there  are  1760  yards  in  a  mile, 

—  number  of  seconds  it  takes  A  to  run  a  mile. 

X 

Since  B  has  a  start  of  20  yards,  he  runs  1740  yards  the  first  trial ; 
and  as  he  was  30  seconds  longer  than  A, 


1  *7f>r) 

+  80  ==  the  number  of  seconds  B  was  running. 

But  =  the  number  of  seconds  B  was  running. 

"  ,.li«  =  l!??  +  30.  (1) 

V  X 


PROBLEMS.  189 

In  the  second  trial  B  runs  1760  -  9^5^  =  1750j\  yards. 

y  X 

From  the  solution  of  equations  (1)  and  (2),  x  =  5j|,  and  y  •■=  b^^. 

Therefore  A  runs  5^f  yards  a  second,  and  B  runs  5^ 
yards  a  second. 

(2)  A  train,  after  travelling  an  hour  from  A  towards  B, 
meets  with  ah  accident  which  detains  it  half  an  hour  ;  after 
which  it  proceeds  at  four-fifths  of  its  usual  rate,  and  arrives 
an  hour  and  a  quarter  late.  If  the  accident  had  happened 
30  miles  farther  on,  the  train  would  have  been  only  an  hour 
late.     Find  the  usual  rate  of  the  train. 

Since  the  train  was  detained  \  an  hour  and  arrived  1\  hours  late, 
the  running  time  was  f  of  an  hour  more  than  usual. 

Let  y  =  number  of  miles  from  A  to  B, 

and  5  a;  =  number  of  miles  the  train  travels  per  hour. 

Then       y —  5a;  =  number  of  miles  the  train  has  to  go  after  the 
accident. 

Hence     ^^-^ =  number  of  hours  required  usually, 

and  ^ — - —  =  number  of  hours  actually  required. 

.•.  y  ~ ^ =  loss  in  hours  of  running  time. 

4a;  6a; 

But  f  —  loss  in  hours  of  running  time. 

.   y  —  5x     y  —  5a;_3 

4a;  5a;        4 


(1) 


If  the  accident  had  happened  30  miles  farther  on,  the  remainder 
of  the  journey  would  have  been  y  — (5x  +  30),  and  the  loss  in  running 
time  would  have  been  |  an  hour. 

.   y-(5a;  +  30)     y-(5a;  +  30).    1   '  ^21 

4a;  5a;  ^2*  ^) 

From  the  solution  of  equations  (1)  and  (2),  x  =  Q,  and  5a;  =  30. 
Therefore  the  usual  rate  of  the  train  is  30  miles  an  hour. 


190  SCHOOL    ALGEBRA. 

38.  Two  men,  A  and  B,  run  a  mile,  and  A  wins  by  2 
seconds.     In  the  second  trial  B  has  a  start  of  18^  yards, 
and  wins  by  1  second.     Find  the  number  of  yards  each  - 
runs  a  second,  and  the  number  of  miles  each  would  run  in 
an  hour. 

39.  In  a  mile  race  A  gives  B  a  start  of  3  seconds,  and  is 
beaten  by  12^  yards.  In  the  second  trial  A  gives  B  a  start 
of  10  yards,  and  the  race  is  a  tie.  Find  the  number  of 
yards  each  runs  a  second.  At  this  rate,  how  many  miles 
could  each  run  in  an  hour  ? 

40.  In  a  mile  race  A  gives  B  a  start  of  44  yards,  and  is 
beaten  by  1  second.     In  a  second  trial  A  gives  B  a  start  of 
6  seconds,  and  beats  him  by  9|-  yards.     Find  the  number  ' 
of  yards  each  runs  a  second. 

41.  An  express  train,  after  travelling  an  hour  from  A 
towards  B,  meets  with  an  accident  which  delays  it  15  min- 
utes. It  afterwards  proceeds  at  two-thirds  its  usual  rate, 
and  arrives  24  minutes  late.  If  the  accident  had  happened 
5  miles  farther  on,  the  train  would  have  been  only  21 
minutes  late.     Find  the  usual  rate  of  the  train. 

42.  A  train,  after  running  2  hours  from  A  towards  B, 
meets  with  an  accident  which  delays  it  20  minutes.  It 
afterwards  proceeds  at  four-fifths  its  usual  rate,  and  arrives 
1  hour  and  40  minutes  late.  If  the  accident  had  happened 
40  miles  nearer  A,  the  train  would  have  been  2  hours  late. 
Find  the  usual  rate  of  the  train. 


43.  A  and  B  can  do  a  piece  of  work  in  2\  days,  A  and 
C  in  3^  days,  B  and  0  in  3f  days.  In  what  time  can  all 
three  together,  and  each  one  separately,  do  the  work  ? 

44.  A  sum  of  money,  at  interest,  amounts  in  8  months 
to  $1488,  and  in  15  months  to  $1530.  Find  the  principal 
and  the  rate  of  interest. 


PROBLEMS.  191 

45.  A  number  is  expressed  by  two  digits,  the  units'  digit 
being  the  larger.  If  the  number  is  divided  by  the  sum  of 
its  digits,  the  quotient  is  4.  If  the  digits  are  reversed  and 
the  resulting  number  is  divided  by  2  more  than  the  differ- 
ence of  the  digits,  the  quotient  is  14.     Find  the  number. 

46.  A  and  B  together  can  dig  a  well  in  10  days.  They 
work  4  days,  and  B  finishes  the  work  in  16  days.  How 
long  would  it  take  each  alone  to  dig  the  well  ? 

47.  The  denominator  of  the  greater  of  two  fractions  is 
20,  and  this  is  the  greater  of  the  two  denominators.  The 
fraction  formed  by  taking  for  a  numerator  the  sum  of  the 
numerators  of  the  two  fractions,  and  for  a  denominator 
the  sum  of  the  denominators,  is  equal  to  f .  The  fraction 
similarly  formed  with  the  difference  of  the  numerators, 
and  of  the  denominators,  is  equal  to  ^.  The  sum  of  the 
numerators  is  twice  the  difference  of  the  denominators. 
Find  the  fractions. 

48.  A  cistern  can  be  filled  in  5  hours  by  two  pipes,  A 
and  B,  together.  Both  are  left  open  for  3  hours  and  45 
minutes,  and  then  A  is  shut,  and  B  takes  3  hours  and  45 
minutes  longer  to  fill  the  cistern.  How  long  would  it  take 
each  pipe  alone  to  fill  the  cistern  ? 

49.  A  man  put  at  interest  $20,000  in  three  sums,  the 
first  at  5  per  cent,  the  second  at  4^  per  cent,  and  the 
third  at  4  per  cent,  receiving  an  income  of  $905  a  year. 
The  sum  at  4|-  per  cent  is  one-third  as  much  as  the  other 
two  sums  together.     Find  the  three  sums. 

50.  An  income  of  $335  a  year  is  obtained  from  two  in- 
vestments, one  in  4-^  per  cent  stock  and  the  other  in  5  per 
cent  stock.  If  the  4-|-  per  cent  stock  should  be  sold  at 
110,  and  the  5  per  cent  at  125,  the  sum  realized  from  both 
stocks  together  would  be  $8300.  How  much  of  each  stock 
is  there  ? 


192  SCHOOL   ALGEBRA. 

51.  A  boy  bought  some  apples  at  3  for  5  cents,  and 
some  at  4  for  5  cents,  paying  for  the  whole  $1.  He  sold 
them  at  2  cents  apiece,  and  cleared  40  cents.  How  many 
of  each  kind  did  he  buy  ? 

52.  Find  the  area  of  a  rectangular  floor,  such  that  if  3 
feet  were  taken  from  the  length  and  3  feet  added  to  the 
breadth,  its  area  would  be  increased  by  6  square  feet,  but 
if  5  feet  were  taken  from  the  breadth  and  3  feet  added  to 
the  length,  its  area  would  be  diminished  by  90  square  feet. 

53.  A  courier  was  sent  from  A  to  B,  a  distance  of  147 
miles.  After  7  hours,  a  second  courier  was  sent  from  A, 
who  overtook  the  first  just  as  he  was  entering  B.  The  time 
required  by  the  first  to  travel  17  miles  added  to  the  time 
required  by  the  second  to  travel  76  miles  is  9  hours  and 
40  minutes.     How  many  miles  did  each  travel  per  hour  ? 

54.  A  box  contains  a  mixture  of  6  quarts  of  oats  and  9 
of  corn,  and  another  box  contains  a  mixture  of  6  quarts  of 
oats  and  2  of  corn.  How  many  quarts  must  be  taken  from 
each  box  in  order  to  have  a  mixture  of  7  quarts,  half  oats 
and  half  corn  ? 

55.  A  train  travelling  30  miles  an  hour  takes  21  minutes 
longer  to  go  from  A  to  B  than  a  train  which  travels  36 
miles  an  hour.     Find  the  distance  from  A  to  B. 

56.  A  man  buys  570  oranges,  some  at  16  for  25  cents, 
and  the  rest  at  18  for  25  cents.  He  sells  them  all  at  the 
rate  of  15  for  25  cents,  and  gains  75  cents.  How  many  of 
each  kind  does  he  buy  ? 

57.  A  and  B  run  a  mile  race.  In  the  first  heat  B 
receives  12  seconds  start,  and  is  beaten  by  44  yards.  In 
the  second  heat  B  receives  165  yards  start,  and  arrives  at 
the  winning  post  10  seconds  before  A.  Find  the  time  in 
which  each  can  run  a  mile. 


PROBLEMS.  193 


Indeterminate  Problems. 

195.  If  a  single  equation  is  given  which  contains  two 
unknown  numbers,  and  no  other  condition  is  imposed,  the 
number  of  its  solutions  is  unlimited;  for,  if  any  value  be 
assigned  to  one  of  the  unknowns,  a  corresponding  value 
may  be  found  for  the  other.  Such  an  equation  is  said  to 
be  indeterminate. 

196,  The  values  of  the  unknown  numbers  in  an  inde- 
terminate equation  are  dependent  upon  each  other ;  so  that, 
though  they  are  unlimited  in  number,  they  are  confined  to 
a  particular  range. 

This  range  may  be  still  further  limited  by  requiring  these 
values  to  satisfy  some  given  condition;  as,  for  instance,  that 
they  shall  be  positive  integers.  With  such  restrictions  the 
equation  may  admit  of  a  definite  number  of  solutions. 

Ex.  A  number  is  expressed  by  two  digits.  If  the  num- 
ber is  divided  by  the  sum  of  its  digits  diminished  by  4,  the 
quotient  is  6.     Find  the  number. 

The  single  statement  is 


x^y 

-4 

Whence 

4a;  =  52/ -24, 

and 

"We  see  from  ^  that  the  values  of  y  which  will  be  integral  are  4,  8, 

12, 16,  or  some  other  multiple  of  4,  and  from  the  relation  a;  =  2/  — 6  +| 

that  the  least  positive  integral  value  of  y  which  will  give  to  x  a,  posi- 
tive integral  value,  is  8.  If  we  put  8  for  y  in  (1),  we  find  a;=»4.  Hence 
the  number  re(juired  is  48, 


194  SCHOOL    ALGEBRA. 


Exercise  66. 

1.  A  number  is  expressed  by  two  digits.  If  the  number 
is  divided  by  the  last  digit,  the  quotient  is  15.  Find  the 
number. 

2.  A  number  is  expressed  by  three  digits.  The  sum  of 
the  digits  is  20.  If  16  is  subtracted  from  the  number  and 
the  remainder  divided  by  2,  the  digits  will  be  reversed. 
Find  the  number. 

Here  x  +y  +  z=^20, 

,                      100a;  +  10y +  2-16     ,.. 
and  ^ =  IOO2  +  lOy  +  x. 

Eliminate  y  and  reduce,  and  we  have 

4a;  =  72  +  8. 

3.  A  man  spends  $114  in  buying  calves  at  $5  apiece, 
and  pigs  at  $  3  apiece.     How  many  did  he  buy  of  each  ? 

4.  In  how  many  ways  can  a  man  pay  a  debt  of  $87 
with  five-dollar  bills  and  two-dollar  bills  ? 

5.  Find  the  smallest  number  which  when  divided  by  5 
or  by  7  gives  4  for  a  remainder. 

Let  n  =  the  number,  then  ^~     =  x,  and ^~    =  y. 
5  7 

6.  A  farmer  sells  15  calves,  14  lambs,  and  13  pigs  for 
$200.  Some  days  after,  at  the  same  price,  he  sells  7  calves, 
11  lambs,  and  16  pigs,  for  which  he  receives  $141.  "What 
was  the  price  of  each  ? 

Discussion  op  Problems. 

197.  The  discussion  of  a  problem  consists  in  making 
various  suppositions  as  to  the  relative  values  of  the  given 
numbers,  and  explaining  the  results.  We  will  illustrate  by 
an  example : 


PROBLEMS.  195 

Two  couriers  were  travelling  along  the  same  road,  and  in 
the  same  direction,  from  C  towards  D.  A  travels  at  the 
rate  of  m  miles  an  hour,  and  B  at  the  rate  of  n  miles  an 
hour.  At  12  o'clock  B  was  d  miles  in  advance  of  A.  When 
will  the  couriers  be  together? 

Suppose  they  will  be  together  x  hours  after  12.  Then  A  has  trav- 
elled mx  miles,  and  B  has  travelled  nx  miles,  and  as  A  has  travelled 
d  miles  more  than  B 

mx  =  nx  +  d, 
or  mx  —  nx='d, 

d 


of  X,  namely,  ,  is  positive,  and  it  is  evident  that  A  will  over- 

m  —  n 


Discussion  op  the  Peoblem.   1.  If  m  is  greater  than  n,  the  value 

X,  namely,   ,  is 

m  —  n 

take  B  after  12  o'clock. 

2.  If  m  is  less  than  n,  then will  be  negative.    In  this  case 

m  —  n 
B  travels  faster  than  A,  and  as  he  is  d  miles  ahead  of  A  at  12  o'clock, 
it  is  evident  that  A  cannot  overtake  B  after  12  o'clock,  but  that  B 

passed  A  hefore  12  o'clock  by hours.    The  supposition,  there- 

n  —  m 
fore,  that  the  couriers  were  together  after  12  o'clock  was  incorrect, 
and  the  negative  value  of  x  points  to  an  error  in  the  supposition. 

3.  If  m  equals  n,  then  the  value  of  x,  that  is, .,  assumes  the 

m  —n 

form  -.     Now  if  the  couriers  were  d  miles  apart  at  12  o'clock,  and  if 

they  had  been  travelling  at  the  same  rates,  and  continue  to  travel  at 
the  same  rates,  it  is  obvious  that  they  never  had  been  together,  and 
that  they  never  will  be  together,  so  that  the  symbol  -  may  be  regarded 
as  the  symbol  of  impossibility. 

4.  If  m  equals  n  and  d  is  0,  then becomes  -.    Now  if  the 

m—n  0 

couriers  were  together  at  12  o'clock,  and  if  they  had  been  travelling 
at  the  same  rates,  and  continue  to  travel  at  the  same  rates,  it  is 
obvious  that  they  had  been  together  all  the  time,  and  that  they  will 
continue  to  be  together  all  the  time,  so  that  the  symbol  -  may  be 
regarded  as  the  symbol  of  indetermination. 


196  SCHOOL    ALGEBRA. 


Exercise  67. 

1.  A  train  travelling  h  miles  per  hour  is  m  hours  in 
advance  of  a  second  train  which  travels  a  miles  per  hour. 
In  how  many  hours  wilj  the  second  train  overtake  the  first  ? 

Am.  -*2L 
a~o 

Discuss  the  result  (1)  when  ay-h\  (2)  when  a  =  6 ;  (3)  when  a <  6. 

2.  A  man  setting  out  on  a  journey  drove  at  the  rate  of 
a  miles  an  hour  to  the  nearest  railway  station,  distant  h 
miles  from  his  house.  On  arriving  at  the  station  he  found 
that  the  train  for  his  place  of  destination  left  c  hours  before. 
At  what  rate  should  he  have  driven  in  order  to  reach  the 
station  just  in  time  for  the  train?  , 

Ans. 


b  —  ac 

Discuss   the  result  (1)  when   c=-0;    (2)  when   c  —  -\   (3)  when 

h  ^ 

c  = .     In  case  (2),  how  many  hours  did  the  man  have  to  drive 

from  his  house  to  the  station?     In  case  (3),  what  is  the  meaning  of 
the  negative  value  of  c  ? 

3.  A  wine  merchant  has  two  kinds  of  wine  which  he  sells, 
one  at  a  dollars,  and  the  other  at  h  dollars  per  gallon.  He 
wishes  to  make  a  mixture  of  I  gallons,  which  shall  cost  him 
on  the  average  m  dollars  a  gallon.  How  many  gallons 
must  he  take  of  each  ? 

Ans.   ^ — "^  /    of  the  first ;  •^^ —^  of  the  second. 

a  —  o  a—  0 

Discuss  the  question  (1)  when  a  =  6 ;  (2)  when  a  or  &  ==  m ;  (3) 
when  a  =  h  =  m;  (4)  when  a>6  and  <m;  (5)  when  a>&  and 
6  >  m. 


CHAPTER  XIII. 

INEQUALITIES. 

198.  An  inequality  consists  of  two  unequal  numbers 
connected  by  the  sign  of  inequality.  Thus,  12  >  4  and 
4  <  12  are  inequalities. 

199.  Two  inequalities  are  said  to  be  of  the  same  direction 
if  the  first  members  are  both  greater  or  both  less  than  the 
second  members ;  that  is,  if  the  signs  of  inequality  point  in 
the  same  direction. 

200.  Two  inequalities  a:re  said  to  be  the  reverse  of  each 
other  if  the  signs  point  in  opposite  directions. 

201.  If  equal  numbers  are  added  to,  or  subtracted  from, 
the  members  of  an  inequality,  the  inequality  remains  in 
the  same  direction.  Thus,  if  a  >  J,  then  a  +  c  >  5  -f-  c,  and 
a  — c>5  —  c.     Hence, 

A  term  can  be  transposed  from  one  member  of  an  in- 
equality to  the  other  without  altering  the  inequality^  provided 
its  sign  is  changed. 

202.  If  unequals  are  taken  from  equals,  the  result  is  an 
inequality  which  is  the  reverse  of  the  given  inequality. 
Thus,  \i x  =  y,  and  a'>b,  then  x  —  a<,y  —  h. 

203.  If  the  signs  of  the  terms  of  an  inequality  are 
changed,  the  inequality  is  reversed.  Thus,  if  a  >  5,  then 
-a<-h.     (See  §33.) 


198  SCHOOL   ALGEBRA. 

204.    Hence,  if  the  members  of  an  inequality  are  multi- 
plied or  divided  by  the  same  positive  number,  the  inequality 
y      remains  in  the  same  direction,  by  the  same  negative  num- 
ber, the  inequality  is  reversed. 

(1)  Simplify  4:r-3>^-| 

We  have  4a;_3>^-^. 

Multiply  by  10,        40  a;  -  30  >  15  a;  -  6. 
Transpose,  25  a;  >  24. 

Divide  by  25,  a;  >  f f . 

Therefore  the  value  of  x  is  greater  than  ^. 

(2)  Find  the  limits  of  a;,  given 

x  —  4:>2  —  Sx, 
Sx  —  2<xi-S. 

We  have  a;-4>2-3a;,  (1) 

and  3a;- 2  <  a; +  3.  (2)  , 

Transpose  in  (1),  4  a;  >  6. 

/.  X  >  IJ. 
Transpose  in  (2),  2  a;  <  5. 

/.  X  <  2J. 

Therefore,  the  value  of  x  lies  between  1^  and  2^. 

(3)  If  a  and  b  stand  for  unequal  and  positive  numbers, 
thena^  +  b^>2ab. 

Since  (a —  6)'  is  positive,  whatever  the  values  of  a  and  b, 
{a-by>0. 
a^-2ab  +  b^>0. 
.'.  a^  +  b^>  2ab. 


INEQUALITIES.  199 


Exercise  68. 

1.  Simplify  (a;+l)^<a;'^-f  3a:-5. 

2.  Simplify  l^-::i^^3-5£, 

3.  Simplify  a: +  25  >  1  x, 

4.  Simplify  3a;-2<-+7J. 

Find  the  limiting  values  of  x,  given 

5.  4a;-6<2a;  +  4, 
2a7+4>16-2a;. 

6. [-hx—ah'>—, 

5  5 

hx  ,     1  ^  ^' 

— —  ax-+-ao  <,—' 

7  7 

Find  the  integral  value  of  x,  given 

7.    J(:r  +  2)  +  ia:<|(a;-4)  +  3, 
i(a;+2)  +  i:r>|(:r  +  l)  +  i 

8.  Twice  a  certain  integral  number  increased  by  7  is 
not  greater  than  19  ;  and  three  times  the  number  dimin- 
ished by  5  is  not  less  than  13.     Find  the  number. 

If  the  letters  stand  for  unequal  and  positive  numbers, 
show  that 

9.  a''  +  W>2h{a  +  h). 

10.  a^  +  ¥>a^h^ab\ 

11.  a?-\-h'^c'>ab  +  ac  +  hc. 

12.  a}h-\-  a^c  +  ah''  +  5V  +  ac"  +  hc'>% ahc. 

13.  ^+^>2.  14.    ^±i>    2«^ 


b     a  2         a-\-b 


CHAPTER  XIV. 

INVOLUTION    AND    EVOLUTION. 

205.  The  operation  of  raising  an  expression  to  any  re- 
quired 'power  is  called  Involution. 

Every  case  of  involution  is  merely  an  example  of  multi- 
plication,  in  which  the  factors  are  equal. 

206.  Index  Law.     If  m  is  a  positive  integer,  by  definition 

a"'  =  aX  aXa to  m  factors. 

Consequently,  if  m  and  n  are  both  positive  integers, 
(a")*"—  a"  X  a"  X  a**  •••••  to  m  factors 

=  {aXa to  n  factors)(a  X  a to  n  factors) 

taken  w  times 

=  a  X  a  X  a to  mn  factors. 

This  is  the  index  law  for  involution. 

207.  Also,  {a!^y=^  a"'"  =  (a^)*". 
And  (ahy=  ab  X  ab to  n  factors 

=  (a  X  a to  n  factors)(5  X  b to  n  factors) 

208.  If  the  exponent  of  the  required  power  is  a  composite 
number,  the  exponent  may  be  resolved  into  prime  factors, 
the  power  denoted  by  one  of  these  factors  found,  and  the 
result  raised  to  a  power  denoted  by  a  second  factor  of  the 
exponent ;  and  so  on.  Thus,  the  fourth  power  may  be  ob- 
tained by  taking  the  second  power  of  the  second  power ; 


INVOLUTION    AND    EVOLUTION.  201 

the  sixth  by  taking  the  second  power  of  the  third  power ; 
and  so  on. 

209.  From  the  Law  of  Signs  in  multiplication  it  is  evi- 
dent that  all  even  powers  of  a  number  are  positive ;  all  odd 
powers  of  a  number  have  the  same  sign  as  the  number  itself. 

Hence,  no  even  power  of  awy  number  can  be  negative; 
and  the  even  powers  of  two  compound  expressions  which 
have  the  same  terms  with  opposite  signs  are  identical. 

Thus,        (b-ay=\-(a-b)l'=^{a-  h)\ 

210.  Binomials.     By  actual  multiplication  we  obtain, 

{a-\-hy=^a^^2ah^h'') 

(a+hj^a^^^a'h  +  Zah''-\-  h'y 

(a  +  by  =a*+4a'b  +  6  a'b'  +  4:ab'  +  b\ 

In  these  results  it  will  be  observed  that : 

I.  The  number  of  terms  is  greater  by  one  than  the  ex- 
ponent of  the  power  to  which  the  binomial  is  raised. 

II.  In  the  first  term,  the  exponent  of  a  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised; 
and  it  decreases  by  one  in  each  succeeding  term. 

III.  b  appears  in  the  second  term  with  1  for  an  exponent, 
and  its  exponent  increases  by  1  in  each  succeeding  term. 

IV.  The  coefficient  of  the  first  term  is  1. 

V.  The  coefficient  of  the  second  term  is  the  same  as  the 
exponent  of  the  power  to  which  the  binomial  is  raised. 

VI.  The  coefficient  of  each  succeeding  term  is  found 
from  the  next  preceding  term  by  multiplying  the  coefficient 
of  that  term  by  the  exponent  of  a,  and  dividing  the  product 
by  a  number  greater  by  one  than  the  exponent  of  b. 

If  5  is  negative,  the  terms  in  which  the  odd  powers  of  b 
occur  are  negative.     Thus, 


202  SCHOOL   ALGEBRA. 

(1)  (a  -  hy  =  a^-Za^h  +  Sab'-h\ 

(2)  (a-by=a*-4:a'b  +  6a'b'  -4:ab'  +  h\ 

By  the  above  rules  any  power  of  a  binomial  of  the  form 
adzb  may  be  written  at  once. 

Note.  The  double  sign  ±  is  read  plus  or  minus ;  and  a±h  means 
a  +  h  or  a  —  h. 

211.  The  same  method  may  be  employed  when  the  terms 
of  a  binomial  have  coefficients  or  exponents. 

Since      (a  -by=a'-d  a'b  +  Sab'-  b\ 
putting  5x^  for  a,  and  2?/'^for  J,  we  have 
(5r'-2y»y, 
=  (5xJ  -  3  (6xy(2f)  +  3  (^x'X^fy  -  {2f)\ 
=  125a;«-  150a;y  +  GOaj^y'-  8y». 

Since      (a  -  5)*  =  a*  -  4  a'5  +  6  a?b^  -  4  a^>'  +  i*, 
putting  ic"  for  a,  and  Jy  for  5,  we  have 

(^-iy)^ 

^a^-2x'y  +  ^xY  -\xY-\-  rVy*- 

212.  In  like  manner,  a  polynomial  of  three  or  more  terms 
may  be  raised  to  any  power  by  enclosing  its  terms  in  paren- 
theses, so  as  to  give  the  expression  the  form  of  a  binomial. 
Thus, 

(1)   (a  +  i  +  c)'  =  [a  +  (^  +  e)]', 

=  a»  + 3a'(5  +  c) -f  3a(6  +  c)»  +  (^  +  c)'. 
=  a»  +  3a'5  +  3aV-f  3a6'+  6a5tf 

+  3ac»4-i'  +  3Z>V  +  8i(?'-f  c'. 


INVOLUTION    AND   EVOLUTION.  203 

(2)   (af^-2x'  +  Sx-{-4:y, 

^(x'-2xy  +  2(ar'~2x')(Sx-i-4:)  +  (Sx  +  4:y, 
=  x^-4:a^+4:x*  +  6x'-4:x^-16x'-\-9x''+2ixi-16, 
=  a;*— 4a;5  +  10^*~4a;'— 7a;'  +  24a:+16. 

Exercise  69. 
Raise  to  the  required  power  : 

1.  {a*y.  11.    (x'-2y. 

2.  (a'b^.  12.    (x  +  Sy. 

3.  f^-^\\  13.    i2x  +  iy. 
\3aby  ^  ^ 

4.  (-5a5V)^  '''    ^'^'-'^'^ 

5.  (-7.^3)3  15.    (2.  +  3y)^ 

.      ,..y^h^n*\^  16.    (2x~7/y. 

6. 


17.    (:ry-2y. 

7.  (-^.yy.  •       ,3^    (l_,  +  ^)>. 

8.  (-3aW/.  ^3^    (l-2.  +  3^)^ 


9. 


(-¥)■ 


20.    (l-a  +  a')'. 


10.    (a:  +  2)^  21.    (3 —  4a;  +  5a;')'. 


Evolution. 

213.  The  nth  root  of  a  number  is  one  of  the  n  equal 
factors  of  that  number. 

The  operation  of  finding  any  required  root  of  an  expres- 
sion is  called  Evolution. 


204  SCHOOL    ALGEBRA. 

Every  case  of  evolution  is  merely  an  example  oi factor- 
ing, in  which  the  required  factors  are  all  equal.    Thus,  the 

square,  cube,  fourth,  roots  of  an  expression  are  found 

by  taking  one  of  its  two,  three,  four equal  factors. 

The  symbol  which  denotes  that  a  square  root  is  to  be 
extracted  is  ->/;  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  written  above  to  indicate  the  root; 
thus,  •{/,  -y/,  etc.,  signifies  the  third  root,  fourth  root,  etc. 

214.  Index  Law.  If  m  and  n  are  positive  integers,  we 
have,  (§  206),  ^^ 

Consequently,  Va^  =  a"". 

Thus,  the  cube  root  of  a^  is  a^ ;  the  fourth  root  of  81  a^^ 
is  3a^ ;  and  so  on. 

This  is  the  index  law  for  evolution. 

215.  Also,  since       (aby  =  a%'^, 
conversely,  ■\/a%'^  =  ab  =  Va"  X  V^, 
and  Vab  =  Va  X  ^b. 

Hence,  to  find  the  root  of  a  simple  expregsion  : 
Divide  the  exponent  of  each  factor  by  the  index  of  the  root, 
and  take  the  product  of  the  resulting  factors. 

216.  From  the  Law  of  Signs  it  is  evident  that 

I.  Any  even  root  of  a  positive  number  will  have  the 
double  sign,  ±. 

II.  There  can  be  no  even  root  of  a  negative  number. 
For  V— ^  is  neither  -{-x  nor  —x',  since  the  square  of 

■\-x  =  -\'X'^,  and  the  square  of  —  x  —  -\-x^. 

The  indicated  even  root  of  a  negative  number  is  called 
an  imaginary  number. 

III.  Any  odd  root  of  a  number  will  have  the  same  sign 
as  the  number. 


INVOLUTION    AND    EVOLUTION. 


205 


Thus. 


116  X"^  4:X  9/ JT= o-i  o  , 


4 


16^, 


2^ 


217.  If  the  root  of  a  number  expressed  in  figures  is  not 
readily  detected,  it  may  be  found  by  resolving  the  number 
into  its  prime  factors.  Thus,  to  find  the  square  root  of 
3.415.104: 


2' 

3415104 

2' 

426888 

3^ 

53361 

7 

5929 

7 

847 

1 

121 

11 


3,415,104  =  2«x  3^  X  7^  X  IP. 


•.  V3,415,104  =  2^x3  x7  Xll  =  1848. 


Simplify : 

1.  V4^*. 

2.  v'ei^. 

3.  </W¥y^\ 


4.  V-32a^ 

5.  V-21x\ 


6.    V25a*. 


7.  </-8a''b\ 

8.  -v^M^, 


Exercise  70, 
9.    ■V~2l6q}\ 

10.  Vm^\ 


11.    V243yV^ 


12.    V-1728c^l 


13.  V-343a^ 

14.  </WJ\ 

15.  ■^b\2a>'%'\ 

16.  -^/Py^. 


17. 


[9^ 
\16:rV 

18.    ^ 


8xY 


21  z' 


19 


20 


21 


32a^« 


243  o;^^ 


■4__ 

4J_16£^ 

*    \ SI  a'b'' 

3/125^ 
\216a^*" 


206  SCHOOL   ALGEBRA. 


Square  Roots  of  Compound  Expressions. 

218.  Since  the  square  of  a~\-b  is  a^-}-2ab-{-b^,  the 
square  root  oi  a^  -\-  2  ab  -{-  b^  is  a  +  b. 

It  is  required  to  find  a  method  of  extracting  the  root 
a-\-b  when  a^-}-2ab  -^P  is  given : 

Ex.   The  first  term,  a,  of  the  root  is  obviously  the  square  root  of 

the  first  term,  a^,  in  the  expression. 

a^  +  2ab  +  b"^ \a  +  b-         If  the  a^  is  subtracted  from  the  given 

a'  expression,   the  remainder  is   2ab  +  b'^. 

2a +  b       2ab  +  b^  Therefore  the  second  term,  b,  of  the  root 

2ab  +  b^  is  obtained  v\rhen  the  first  term  of  this 

remainder  is  divided  by  2a,  that  is,  by 

double  the  part  of  the  root  already  found.     Also,  since 

2ab  +  b^=={2a  +  b)b, 

the  divisor  is  completed  by  adding  to  the  trial-divisor  the  new  term  of 
the  root. 

(1)   Find  the  square  root  of  2bx^  —  20x^y-\~^x'^y'^. 
25a;2  -  203^y  +  4:X*y^\5x-2x^y 


25x2 
10x-2x''y 


-203^y  +  4:xY 
-20a^y +  4a;V 


The  expression  is  arranged  according"  to  the  ascending  powers  of  x. 

The  square  root  of  the  first  term  is  5  a;,  and  5  a;  is  placed  at  the 
right  of  the  given  expression,  for  the  first  term  of  the  root. 

The  second  term  of  the  root,  —2x^y,  is  obtained  by  dividing 
—  20a^y  by  10  a;,  and  this  new  term  of  the  root  is  also  annexed  to  the 
divisor,  10  a;,  to  complete  the  divisor. 

219.  The  same  method  will  apply  to  longer  expressions, 
if  care  be  taken  to  obtain  the  trial-divisor  at  each  stage  of 
the  process,  by  doubling  the  part  of  the  root  already  found, 
and  to  obtain  the  complete  divisor  by  annexing  the  new  term 
of  the  root  to  the  trial-divisor. 


INVOLUTION   AND    EVOLUTION.  207 

Ex.    Find  the  square  root  of 

16a;g-24a;5^25a;^-20a^+10a;^-4a;  +  l|4a;«-3a;g  +  2a;-l 
16  x^ 


8a^ 


-3x^ 

-24x5  + 
-24^5  + 

25  x^ 
dx* 

-20x3  +  10x« 
-12x3+   43,2 

Sa^- 

6x2  + 2a; 

16  X* 
\6x* 

8a;'-6a;2  +  4a 

;-l 

-  8x3+    6x2-4x  +  l 

-  8x3+    6x2-4x  +  l 

The  expression  is  arranged  according  to  the  descending  powers  of  x. 
It  will  be  noticed  that  each  successive  trial-divisor  may  be  obtained 
by  taking  the  preceding  complete  divisor  with  its  last  term  doubled. 


Exercise  71. 
Find  the  square  root  of 

1.  x^-Sx^+lSx'-Sx+l, 

2.  9a*-6a'-f-13a^-4a  +  4. 

3.  4a;*-12a:V  +  29a:y-30a;3/'  +  253^. 

4.  l  +  4:x+10x^  +  12x'  +  9x\ 

5.  16 -96:^  +  216:^2-2160;^+ 81a;*. 

6.  x'-22x'i-9bx'  +  286x+169. 

7.  4:X*~llx^-{-25~12x^  +  d0x. 

8.  9a;*  +  49-12a;'— 28:r  +  46a;l 

9.  49a;* +  126r''+ 121 -73a;' -198a;. 
10.    16a;*-30a;-31a;2  +  24a;-'  +  25. 

a;*      2a;'  ,  3a;'      2a; 


11.   r__r:r.+  ^ 
a*       a'        a* 


+  1. 


208 


SCHOOL    ALGEBRA. 

12. 

4:X' 

'  +  4^     i-+,V 

13. 

'+«  +  f- 

14. 

a'- 

-2.3  +  3^ 

2+1. 

15.  ^.  +  2^+10^'  +  2+^- 

3  9        3     4 

16.  4^+3^  +  41  + 3^+ J^. 

y^        y       16     4^7     4  a:'* 

17.  ^~ax  +  ^  +  x^-^x-\--- 
4  ,2  4 

18.  16^*  +  J^a;V  +  8a;^  +  f/  +  Jy  +  l. 
-_     9a;*     3a;'  ,  43:r'      Ta;  ,  49 

20.    4a''  +  -^---ll  +  4a. 


Find  to  three  terms  the  square  root  of 

21.  a^-{-h.               24.    1  +  a.  27.  4a;' +  3. 

22.  a;'H-Jy.            25.    l-2a.  28.  4 -3a. 

23.  l  +  2a.             26.    4a' +  25.  29.  4a2-l. 

220.  Arithmetical  Square  Eoots.  In  the  general  method 
of  extracting  the  square  root  of  a  number  expressed  by 
figures,  the  first  step  is  to  mark  off  the  figures  in  groups. 

Since  1  =  P,  100  =  10',  10,000  =  100',  and  so  on,  it  is 
evident  that  the  square  root  of  any  number  between  1  and 
100  lies  between  1  and  10  ;  the  square  root  of  any  number 
between  100  and  10,000  lies  between  10  and   100.     In 


INVOLUTION   AND   EVOLUTION.  209 

other  words,  the  square  root  of  any  number  expressed  by- 
owe  or  two  figures  is  a  number  of  one  figure  ;  the  square  root 
of  any  number  expressed  by  three  or  four  figures  is  a  num- 
ber of  two  figures ;  and  so  on. 

If,  therefore,  an  integral  square  number  is  divided  into 
groups  of  two  figures  each,  from  the  right  to  the  left,  the 
number  of  figures  in  the  root  will  be  equal  to  the  number 
of  groups  of  figures.  The  last  group  to  the  left  may  have 
only  one  figure. 

Ex.    Find  the  square  root  of  3249. 

32  49(57  In  this  case,  a  in  the  typical  form  a^  +  2ah  -^h"^ 

25  represents  5  tens,  that  is,  50,  and  b   represents  7. 

107)  749  The  25  subtracted  is  really  2500,  that  is,  a^  and  the 

749  complete  divisor  2a+6is2x50+7  =  107. 

221.  The  same  method  will  apply  to  numbers  of  more 
than  two  groups  of  figures  by  considering  a  in  the  typical 
form  to  represent  at  each  step  the  part  of  the  root  already 
found. 

It  must  be  observed  that  a  represents  so  many  tens  with 
respect  to  the  next  figure  of  the  root. 

Ex.   Find  the  square  root  of  5,322,249. 

5  32  22  49(2307 

4 
43) 132 

129 
4607)32249 
32249 

222»  If  the  square  root  of  a  number  has  decimal  places, 
the  number  itself  will  have  twice  as  many.  Thus,  if  0.21  is 
the  square  root  of  some  number,  this  number  will  be  (0.21)^ 
=  0.21  X  0.21  =  0.0441 ;  and  if  0.111  be  the  root,  the  num- 
ber will  be  (0.111)^  -  0.111  X  0.111  =  0.012321. 


210  SCHOOL    ALGEBRA. 

Therefore,  the  number  of  decimal  places  in  every  square 
decimal  will  be  even,  and  the  number  of  decimal  places  in 
the  root  will  be  half  as  many  as  in  the  given  number  itself. 

Hence,  if  a  given  number  contains  a  decimal,  we  divide 
it  into  groups  of  two  figures  each,  by  beginning  at  the 
decimal  point  and  marking  toward  the  left  for  the  integral 
number,  and  toward  the  right  for  the  decimal.  "We  must 
be  careful  to  have  the  last  group  on  the  right  of  the  deci- 
mal point  contain  two  figures,  annexing  a  cipher  when 
necessary. 

Ex.    Find  the  square  roots  of  41.2164  and  965.9664. 

41.21  64  (6.42  9  65.96  64  (31.08 

36  9 

124)521  61)^ 

496  61 


1282)2564  6208)49664 

2564  49664 

223.  If  a  number  contains  an  odd  number  of  decimal 
places,  or  if  any  number  gives  a  remainder  when  as  many 
figures  in  the  root  have  been  obtained  as  the  given  number 
has  groups,  then  its  exact  square  root  cannot  be  found.  We 
may,  however,  approximate  to  its  exact  root  as  near  as  we 
please  by  annexing  ciphers  and  continuing  the  operation. 

The  square  root  of  a  common  fraction  whose  denominator 
is  not  a  perfect  square  can  be  found  approximately  by 
reducing  the  fraction  to  a  decimal  and  then  extracting  the 
root ;  or  by  reducing  the  fraction  to  an  equivalent  fraction 
whose  denominator  is  a  perfect  square,  and  extracting  the 
square  root  of  both  terms  of  the  fraction.     Thus, 

|5 


^l 


0.79057 : 


or        ^(-  =  ^|-  =  VI0  =  V10_316227_0^7905^_ 


>li=>l 


Vl6 


INVOLUTION    AND   EVOLUTION. 


211 


Ex.   Find  the  square  roots  of  3  and  357.357. 


3.(1.732 

1 

27)200 
189 

3  57.35  70  (18.903 

1 
28) 257 
224 

343) 1100 
1029 

369) 3335 
3321 

3462)7100 
6924 

37803) 147000 
113409 

"Rxercise  72. 

- 

Find  the  square  root  of 

1.    289.                  6. 

150.0625. 

11. 

640.343025. 

2.    1225.                 7. 

118.1569. 

12. 

100.240144. 

3.    12544.              8. 

172.3969. 

13. 

316.021729. 

4.    253009.             9. 

5200.140544. 

14. 

454.585041. 

6.    529984.           10. 

1303.282201. 

15. 

5127.276025. 

Find  to  four  decimal  places  the  square  root  of 

16.    10.         19.    0.5. 

22.    0.607. 

25. 

I        28.    f 

17.    3.           20.    0.7. 

23.    0.521. 

26. 

|.         29.    f 

18.    5.  21.    0.9.         24.    0.687.         27.    f.        30.    -& 


ti- 


224.   Cube  Boots  of  Oompound  Expressions.     Since  the  cuhe 
oi  a-\-b  is  a^+  3a^6  +  3a6^+  b^,  the  cube  root  of 

a'+Sa'bi-Sab'+b^  is  a-\-b. 

It  is  required  to  devise  a  method  for  extracting  the  cube 
root  a-}-b  when  a'  +  3  a'5  +  3  ab^  +  ^'  is  given  : 


212  SCHOOL    ALGEBRA. 

(1)    Find  the  cube  root  of  aH  3  a^h  +  3  a6'  +  h^. 


3  a' 


+  3a6  +  62 


3a2+3a6  +  62 


3a26  +  3aJ2  +  53 


The  first  term  a  of  the  root  is  obviously  the  cube  root  of  the  first 
term  a?  of  the  given  expression. 

If  G?  is  subtracted,  the  remainder  is  Zc?h  +  3a&'^  +  6^;  therefore, 
the  second  term  h  of  the  root  is  obtained  by  dividing  the  first  term 
of  this  remainder  by  three  times  the  square  of  a. 

Also,  since  3  a^S  +  3  ab"^  +  ^3  =  (3  a^  +  3 ah  +  b^)  b,  the  complete 
divisor  is  obtained  by  adding  3ab  +  b^  to  the  trial-divisor  3  a'. 

(2)   Find  the  cube  root  of  8x'  +  SQx'i/  -j-  54:X^'  i-  27  f. 

8  a;3  +  36  x'^y  +  54  xy^  +  27  y^  \2x  +  Sy 
12x2  8x3 


(6a;  +  3y)3y=        .      ISxy-hdy^ 


12x^  +  18xy  +  9y^ 


36x^y  +  54:xy^  +  27y^ 
3Qx'y  +  5ixy^+21y^ 


The  cube  root  of  the  first  term  is  2x,  and  this  is  therefore  the  first 
term  of  the  root.     8  a^,  the  cube  of  2  x,  is  subtracted. 

The  second  term  of  the  root,  3  y,  is  obtained  by  dividing  36  x'^y  by 
3  (2a;)2=  12a;2,  which  corresponds  to  3a2in  the  typical  form,  and 
the  divisor  is  completed  by  annexing  to  12  x^  the  expression 

{3(20?)  +  3y}Sy  =  I8xy  +  9y\ 

which  corresponds  to  3  a6  +  b^  in  the  typical  form. 

225.  The  same  method  may  be  applied  to  longer  expres- 
sions by  considering  a  in  the  typical  form  Sa^-{-Sab-{-  b'^ 
to  represent  at  each  stage  of  the  process  the  part  of  the  root 
already/  found.  Thus,  if  the  part  of  the  root  already  found 
is  07  +  2/,  then  3  a^  of  the  typical  form  will  be  represented 
by  3(a;  +  3/)2;  and  if  the  third  term  of  the  root  is  -\-z,  the 
Sab-\-b^  will  be  represented  by  3  (^  +  y)  z -f- z^  So  that 
the  complete  divisor,  Za^+Sab-}-  b\  will  be  represented  by 
3(^  +  y)^+3(a:4-3/)2  +  z^ 


INVOLUTION   AND   EVOLUTION. 


213 


Find  the  cube  root  of  x^  —  Sa;^-\-5s^—Sx  —  l. 
\x^  —  X  —  1 


3a^ 


{Sx^-x){-x) 


-3a^+    x^ 


Sx*-'Sx»+    x^ 


x^-3x^  +  5a^-3x-l 
^3x5  +3x^-    ar* 


3(a;2-a;)2  =  3a;*-6a^  +  3; 


{3x^-3x-l){-l) 


3a;2  +Sx  +1 


3a;*-.  6x3 


+  3a;  +1 


3aj*+6x3_3a._i 


3a;*  +  6ar»-3a;-l 


The  root  is  placed  above  the  given  expression  for  convenience  of 
arrangement. 

The  first  term  of  the  root,  x^,  is  obtained  by  taking  the  cube  root 
of  the  first  term  of  the  given  expression ;  and  the  first  trial-divisor, 
3  a;*,  is  obtained  by  taking  three  times  the  square  of  this  term. 

The  first  complete  divisor  is  found  by  annexing  to  the  trial-divisor 
(3 x^  —  x) (—  x)*  which  expression  corresponds  to  (3a  +  b)b  in  the 
typical  form. 

The  part  of  the  root  already  found  (a)  is  now  represented  by  a;^  —  a; ; 
therefore  3  a^  is  represented  by  3  (x'^  —  a;)''  =  3  a;*  —  6  fl::^  4-  3  a;',  the  second 
trial-divisor ;  and  {3a  +  b)b  by  (3  a;^  —  3  a;  —  1)  (—  1) ;  therefore,  in  the 
second  complete  divisor,  3  a'^  -|-  (3  a  -h  &)  6  is  represented  by 

{3x^-ex^  +  3x'')  +  {3  x^  -  3 x  -  1)  X  {-  1)  =  3 x^  -  6a^  +  3 X  +  I. 


Exercise  73. 
Find  the  mbe  root  of 

1.  a^-i-Sa'x+Sax'  +  a^. 

2.  8+12x-i-6x''  +  x\ 

3.  x^-Sax'+ba'x'~Ba'x-a^ 

4.  l-6x-{-21x'-Ux'-{-6Sx*~5^a^-{-27x\ 

5.  l-Sx+6x^-1x'+6x'-Sx^i-x^. 

6.  x^+l-6x-6x'-j-lbx'-^lbx*~20x',    . 


214  SCHOOL    ALGEBRA. 

7.  64:X^~U4:x'-\-8-S6x-\-102x'-l7lx'+204.x\ 

8.  ■27a«-27a^-18a*+17a^+6a^-3a-l. 

9.  8x^-S6a^+66x'-63x'-\-S3x'-9x+l. 
10.    27  +  108a;+90:r^-80a;^-60a:*  +  48:r^-8^^ 

3       27 

226.  Aritlimetical  Cube  Eoots.  In  extracting  the  cube  root 
of  a  number  expressed  by  figures,  the  first  step  is  to 
separate  it  into  groups. 

Since  1  -  P.  1000=  10^  1,000,000  =  100^  and  so  on,  it 
follows  that  the  cube  root  of  any  number  between  1  and 
1000,  that  is,  of  any  number  which  has  one,  two,  or  three 
figures,  is  a  number  of  one  figure ;  and  that  the  cube  root 
of  any  number  between  1000  and  1,000,000,  that  is,  of  any 
number  which  has  four,  five,  or  six  figures,  is  a  number  of 
two  figures  ;  and  so  on. 

If,  therefore,  an  integral  cube  number  be  divided  jnto 
groups  of  three  figures  each,  from  right  to  left,  the  number 
of  figures  in  the  root  will  be  equal  to  the  number  of  groups. 
The  last  group  to  the  left  may  consist  of  one,  two,  or  three 
figures. 

227.  If  the  cube  root  of  a  number  have  decimal  places, 
the  number  itself  will  have  three  times  as  many.  Thus,  if 
0.11  be  the  cube  root  of  a  number,  the  number  is  0.11  X  0.11 
X  0.11  =  0.001331.  Hence,  if  a  given  number  contains  a 
decimal,  we  divide  the  figures  of  the  number  into  groups 
of  three  figures  each,  by  beginning  at  the  decimal  point 
and  marking  toward  the  left  for  the  integral  number,  and 


INVOLUTION   AND   EVOLUTION. 


215 


toward  the  right  for  the  decimal.  We  must  be  careful  to 
have  the  last  group  on  the  right  of  the  decimal  point  con- 
tain three  figures,  annexing  ciphers  when  necessary. 

228.    Notice  that  if  a  denotes  the  first  term,  and  h  the 
second  term  of  the  root,  the  j?rs^  co7nplete  divisor  is 

and  the  second  trial-divisor  is  3  (a  +  ^)^  that  is, 
3a''+6a5  +  35^ 

which  may  be  obtained  by  adding  to  the  preceding  complete 
divisor  its  second  term  and  twice  its  third  term. 

Ex.  Extract  the  cube  root  of  5  to  five  places  of  decimals. 


5.000  (1'.70997 
1 


300 

4000 

210 

49. 
559  \ 

3913 

259  J 

87000000 

8670000 

45900 

81  1 

871^981  i 

78443829 

45981  J 

85561710 

8762043 

78858387 

67033230 

61334301 

3x102 
3(10x7) 


3  X 17002 

3(1700x9) 

92 


3  x  17092 


After  the  first  two  figures  of  the  root  are  found,  the  next  trial- 
divisor  is  obtained  by  bringing  down  the  sum  of  the  210  and  49 
obtained  in  completing  the  preceding  divisor ;  then  adding  the  three 
lines  connected  by  the  brace,  and  annexing  two  ciphers  to  the  result. 

The  last  two  figures  of  the  root  are  found  by  division.  The  rule 
in  such  cases  is,  that  two  less  than  the  number  of  figures  already 
obtained  may  be  found  without  error  by  division,  the  divisor  being 
three  times  the  square  of  the  part  of  the  root  already  found. 


216  SCHOOL    ALGEBRA. 

Exercise  74. 
Find  the  cube  root  of 

1.  4913.  3.    1404928.  5.   385828.352. 

2.  42875.  4.    127263527.  6.    1838.265625. 

Find  to  four  decimal  places  the  cube  root  of 

7.  87.  9.    3.02.         11.   0.05.  13.  |.         15.  ^. 

8.  10.         10.    2.05.         12.   0.677.         14.  f.         16.  ^V 

229.  Since  the  fourth  power  is  the  square  of  the  square, 
and  the  sixth  power  the  square  of  the  cube,  the  fourth  root 
is  the  square  root  of  the  square^  root,  and  the  sixth  root  is 
the  cube  root  of  the  square  root.  In  like  manner,  the 
eighth,  ninth,  twelfth, roots  may  be  found. 

Exercise  75. 
Find  the  fourth  root  of 

1.  Slx'+108x'-{-54:x^-{-12x'^l. 

2.  16x'-S2ax^  +  24:a''x'-8a'x-^a\ 

3.  1  -{-  4:x  +  x^  +  4:x'' -{-10x^+16x'  +  10x'+l9x'+16a^. 

Find  the  sixth  root  of 

5.  729-1458:i;+1215a;^-540a;^  +  135:r-'-18:r^+a;«. 

6.  1  -  18y  + 135^  -  540/  +  1215?/*  -  1458y^  +  729/. 


CHAPTER  XV. 
THEORY  OF  EXPONENTS. 

230,  If  n  is  a  positive  integer,  we  have  defined  a"  to 
mean  the  product  obtained  by  taking  a  as  a  factor  n  times. 
Thus  d?  stands  for  a  X  a  X  a ;   5*  stands  for  hxhxhxh. 

231.  From  this  definition  we  have  obtained  the  following 
laws  for  positive  and  integral  exponents  : 

I.    a"*  X  a"*  =  a"*+". 

II.        («*")"=-«"»". 

III.  -"  =  a"*-",  if  m  >  w. 


IV. 


a" 


V.        {aby^oJ"}/'. 

232.  Since  by  the  definition  of  a"  the  exponent  n  denotes 
simply  repetitions  of  a  as  a  factor,  such  expressions  as  a^  and 
a~^  have  no  meaning  whatever.  It  is  found  convenient, 
however,  to  extend  the  meaning  of  a**  so  as  to  include 
fractional  and  negative  values  of  n. 

233.  If  we  do  not  define  the  meaning  of  a"  when  n  is 
a  fraction  or  negative,  but  require  that  the  meaning  of  a** 
must  in  all  cases  be  such  that  the  fundamental  index  law 
shall  always  hold  true,  namely, 

««  X  a**  =  a"^", 

we  shall  find  that  this  condition  alone  will  be  sufiicient  to 
define  the  meaning  of  a**  for  all  cases. 


218  SCHOOL   ALGEBRA. 

234.   To  find  the  Meaning  of  a  Fractional  Exponent. 

Assuming  the  index  law  to  hold  true  for  fractional  expo- 
nents, we  have 

1  1  1  1  4.  1  4-1  3 

a^Xa^  X  a^  =  d^^^^ ^  =  a^  =  a, 

I  X  11  n 

a«  X  a« to  n  factors  =  a«    «' 


mn 


~     ,      —  ,  c      I  — I — to  n  terms 

a»  X  a» to  71  factors  =  a«    «  =a^=a  . 

That  is,    a^  is  one  of  the  two  equal  factors  of  a, 

a^  is  one  of  the  three  equal  factors  of  a, 

a^  is  one  of  the  four  equal  factors  of  a', 

1 
an  is  one  of  the  n  equal  factors  of  a, 

an  is  one  of  the  n  equal  factors  of  a™. 
Hence      a^  =  Va ;      a^  =  Va ; 

a^=</^';     a«  =  -v^.  (§213) 

1        1        i_ 
Also,    a^x  a»X  a^x to  m  factors.     * 

-  H h  - to  m  terms  - 

m 

.*.  a»  = 

The  meaning,  therefore,  of  a»,  where  m  and  w  are  posi- 
tive integers,  is,  the  wth  root  of  the  mth  power  of  a,  or  the 
mth  power  of  the  wth  root  of  a. 

Hence  the  numerator  of  a  fractional  exponent  indicates 
a  power,  and  the  denominator  a  root ;  and  the  result  is  the 
same  when  we  first  extract  the  root  and  raise  this  root  to 
the  required  power,  as  when  we  first  find  the  power  and 
extract  the  required  root  of  this  power. 


THEORY   OF   EXPONENTS.  219 

235.  To  find  the  Meaning  of  a°. 
By  the  index  law, 

a^Xor  =  a®-*-"*  =  or. 

,    .'.  a^=  1,  whatev^the  value  of  a  is. 

236.  To  find  the  Meaning  of  a  Negative  Exponent. 

If  n  stands  for  a  positive  integer,  or  a  positive  fraction, 
we  have  by  the  index  law, 


But 


That  is,  a"  and  a~"  are  reciprocals  of  each  other  (§  167), 
so  that  a~"  =  — ,  and  a"  = 


237.  Hence,  we  can  change  any /ac^or  from  the  numerator 
of  a  fraction  to  the  denominator,  or  from  the  denominator 
to  the  numerator,  ^rovzc?ec?  we  change  the  sign  of  its  exponent. 

Thus  — —    may  be  written  alfc~^d~^,  or 


a" 

X  a""  = 

-1. 

-n 

— 

a\ 

,-.  a" 

Xa-"  = 

-1. 

238.  We  have  now  assigned  definite  meanings  to  frac- 
tional and  negative  exponents,  meanings  obtained  by 
subjecting  them  to  the  fundamental  index  law  of  positive 
integral  exponents;  and  we  will  now  show  that  Law  II., 
namely,  («"*)"—«"***,  which  has  been  established  for  positive 
integral  exponents,  holds  true  for  fractional  and  negative 
exponents. 

(1)   If  n  is  a  positive  integer,  whatever  the  value  of  m, 

We  have  (a"*)"  =  a*^xa^xa'^ to  n  factors, 

_  ^t»  +  m  +  f» to  n  termi 


220  SCHOOL   ALGEBRA. 

(2)   If  w  is  a  positive  fraction  -L,  where  p  and  q  are  posi- 
tive integers,  we  have  ^ 

p         

{a^)n  =  (a"»)?  =  Via'^y  g  234 

=  V^  (1) 

=  aV  §234 


(3)  If  w  is  a  negative  integer,  and  equal  to  —p,  we  have 

{a^Y  =  (a™)  -p  =  --\-  I  236 

-^  •       (1) 

=  a-^p  I  236 

=  a^{-p) 

(4)  If  n  is  negative  and  equal  to  the  fraction  — -,  where 
p  and  q  are  positive  integers,  we  have 

(a«)»  =  (««)-?= -i_  §236 


1 


1 


§234 
§234 


a  9 
1 


a       9 


1 

a««.  §  236 


(1) 


Hence,  (cry  =  od^'^,  for  a/^  values  of  m  and  w. 


THEORY   OF   EXPONENTS.  221 

239.  In  like  manner  it  may  be  shown  that  all  the  index 
laws  of  positive  integral  exponents  apply  also  to  fractional, 
and  negative,  exponents.    We  will  now  give  some  examples. 

(2)  16*  =  (-v/16)^=-±2'=±8. 

(3)  Jxa'^-=J~^  =  a^''^=Va. 
,(4)    J'xJxa~^  =  J'^^~^  =  a''=l. 

(5)  (a~^y=a'^-^^  =  a-'  =  -^' 

(6)  ^=a-3-(-5)  =  a-3+5_^2^ 

CL 

(7)  Vd'b-'c-'d=aH-'c-^d^. 

(8)  (4a-V^=— ^-  =  -^-— ^  =  -- 


I6.a-V^     rSlb'\^     21b^     21a'b^ 


r9^   /16^-Y^-,/8U^Y_275^ 
^  ^   \8lb'J         \16a-y       8a~' 

(10)   (3^a-')-^ 


(SV')^  .3^"V^     3^     a/3 

Exercise  76. 
Express  with  fractional  exponents : 
1.   Va'.        3.   ■Va'.        5.   </^\        7.   Va+^^+-V^16^. 

2.  V^^     4.  -s/^^.  6.  ^/a^     8.  -y^+-v/^. 


222  SCHOOL   ALGEBRA. 

Express  with  root-signs : 
9.  a^.         11.  a^bK  13.  x^y~K  15.  a^  —  xhK 

10.  oK         12.  aH\         14.  Sx^y'K         16.  a^-\-x^c^. 

Express  with  positive  exponents : 

17.  a-\       19.  Sx-'fr    21.  4a;-'y-^        23.     "^^''^ 

18.  a~i      20.  4a:y-\        22.  Sa'^b^." 


S-'b'y 


Write  without  denominators : 

24.    ^^.  26.         ^^"      .  28.    ^I^. 

a;-^  a-^bc-^d  a-^b~^ 

25        ^'^  27     ^"'^"'^~'  29     ^"'^''^"' 

*    rr-^z-^'  '    a-'b-'c-'  ■    '   a-'b-'c-' 

Find  the  value  of 

30.  8^         32.  27^  34.  36^.  36.  (-27)^x25^ 

31.  16~i     33.  (-8)"i     35.  (-27)i    37.  81"^xl6^. 

Simplify : 

38.  8^X4"^.  40.   (3^)-^x(^V)^.      42.   (a-hY^\ 

39.  (^)2xl6~4.    41.  a'h^Xah^.  43.   (a"^^)-^ 

If  a  =  4,  5  =  2,  c  =  1,  find  the  value  of 

44.  ah~\       46.    a~h\         48.    3(a5)^.  50.   (ab'c)^. 

45.  a^>-^         47.    a~K~^.      49.    2(a5)~i       51.  (ai'c)"*. 


THEORY    OF   EXPONENTS.  223 

240.   Compound  expressions  are  multiplied  and  divided  as 
follows : 


(1)  Multiply  x^  +  x^y^  +  y^  by  x^-x^y^-^yK 
x^  —  x'fy*  +  y^ 


X  +  x^y'f  +  x^y^ 

—  x'3/*  —  x^y^  —  a;*2/' 

+  a;'2/5  +  x^y^  +  y 


+  x^y^ 


+  2/ 


(2)   Divide  ^  +  \/:t;  -  12  by   Vx-  3. 

a;t+    x^-l2\x^-2, 
x^  —  Zx^  a;*  +  4 

+  4a;»-12 
+  4x^-12 


Exercise  77. 
Multiply : 

1.  a^-\-h^   by  a^ —  h^.  3.  a^  —  ^>^  by  a*-ii 

2.  a^4-6^  by  a*  +  5^      4.  x^'\-2x   by  a;^-2a:. 
5.  a;~'^4-a;~^2/~^  +  y~^  by  a;"^  —  a:~^2/~^  +  2/~'^• 
6.  x^  —  x^y^-\-y^   by  a;*  +  y*. 

7.  x^  ■{■  x^ y^ -\- y^   by  x^  —  y^. 

8.  i-j-J-i  +  i-'  by  1-5-H^-'- 

9.  a7*  +  a;*y^  +  y*  by  x^  —  x^y^-^y^. 

10.  a^6"^  +  2a^-35^  by  26"^- 4a~*-6a~^^>*. 


224  SCHOOL   ALGEBRA. 

Divide : 

11.  a-b  by  J  —  b^.  13.    a-b  by  a^  —  b^. 

12.  a  +  b  by  a^  +  Z>i  14.    a  +  J  by  a^  +  ii 

15.  2a:-'  +  6a;-V-'  — lea^V"  by  2a:  +  2a:V~^  +  4a:'y-'. 

16.  a:  +  y  +  2  —  3a:* y^ 2^  by  a;*  +  y*  +  z* 

17.  a:  — 3a;^  +  3a;^-l  by  a;*-l. 

18.  x  +  x^y^  +  7/  by  a;^  — a:^y^  +  yi 

19.  a;^-4a;^  +  l  +  6a;~^  by  a;*-2. 

20.  9a7-12a;^-2  +  4a;~^  +  a:-^  by  Sx^-2-x-K 
Find  the  square  root  of 

21.  a;^+2a:^+l.  23.    x^-4:X^  +  4:. 

22.  4:J~4:ah^-i-b^.  24.    4a-''  +  4a-^+l. 

25.  9a-12a^+10  — 4a"^4-a-\ 

26.  49a;3-28a:+18a:'-4a:^+l. 

27.  ^^^  +  2771-1-2^-^  +  ^-'. 

28.  l+4y"^~2y"^-4y-^+25y"^-24y~3  +  16y-^ 

Expand : 

29.  (a:-a?-^)».       31.  (2a^- a-^)l     33.  (iV^-J-v^)*. 

30.  (v^-^a:)\    32.  (2a;-=^  +  a;^/.      34.  (iV^-^+ia;-^. 


u 


CHAPTER  XVI. 
RADICAL   EXPRESSIONS. 

241.  A  radical  expression  is  an  expression  affected  with 
the  radical  sign ;  as,  V«,  \/9,  Va^,  Va  +  h,  V32. 

242.  An  indicated  root  that  cannot  be  exactly  obtained 
is  called  a  surd.  An  indicated  root  that  can  be  exactly 
obtained  is  said  to  have  the /orw  of  a  surd. 

The  required  root  shows  the  order  of  a  surd ;  and  surds 
are  named  quadratic,  cubic,  biquadratic  .  .  .  according 
as  the  second,  third,  fourth  .  .  .  roots  are  required. 

The  product  of  a  rational  factor  and  a  surd  factor  is 
called  a  mixed  surd;  as,  3V2,  5Va.  The  rational  factor 
of  a  mixed  surd  is  called  the  coefficient  of  the  radical. 

When  there  is  no  rational  factor  outside  of  the  radical 
sign,  that  is,  when  the  coefficient  is  1,  the  surd  is  said  to  be 
entire;  as,  V2,  Va. 

243.  A  surd  is  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  integral  and  as  small  as  possible. 

Surds  which,  when  reduced  to  the  simplest  form,  have 
the  same  surd  factor,  are  said  to  be  similar. 

Note,  In  operations  with  surds,  arithmetical  numbers  contained 
in  the  surds  should  be  expressed  in  their  prime  factors. 

Reduction  of  Radicals. 

244.  To  reduce  a  radical  is  to  change  its  form  without 
changing  its  value. 


226  SCHOOL   ALGEBRA. 

Case  I. 

245.  Wlien  the  radical  is  a  perfect  power  and  has  for  an 
exponent  a  factor  of  the  index  of  the  root. 

(1)  v^  =  al  =  al  =  Va; 

(2)  ■</36^^=-v^^6^=(6a5)f  =  (6a^»)^  =  V6^; 

(3)  A/25a*5V  =  -\/{ba'bcJ  =  (5 a'bc') ^  =  (5 a'bc')^ 

We  have,  therefore,  the  following  rule : 

Divide  the  exponent  of  the  power  by  the  index  of  the  root. 


Simplify : 

Exercise  78. 

1.    </25. 

6.    </^\ 

2.    \/l6. 

7.    </a^b\ 

3.    \/27. 

8.    ■v^a*^^*. 

4.    -v/i9. 

9.    ■y/21a'b\ 

5.    ^64. 

10.    Vl6a*6*. 

11. 


12. 


13. 


25^ 
646^ 


\(x-sy 


Case  II. 


246.  When  the  radical  is  the  product  of  two  factors,  one  of 
which  is  a  perfect  power  of  the  same  degree  as  the  radical. 

Since  VM  =  V^xVb  =  a\/'b{^  215),  we  have 

(1)  V^  =  V^'xVb  =  aVb; 

(2)  Vm=</2f^<^==^s/27x-Vi==SVi\ 


RADICAL   EXPRESSIONS. 


227 


(3)  4V72a^-4V'36a^6'x26  =  4V36a''6^X  V26 

=  4  X  6a^>V26  =  24ai  V2^ ; 

(4)  2\/54^=  2^/27a'  X  2ab  =  2^27^^ x  ^2^ 

=  2xSa</2^  =  6a</2ab. 

We  have,  therefore,  the  following  rule : 

Hesolve  the  radical  into  two  factors,  one  of  which  is  the 
greatest  perfect  power  of  the  same  degree  as  the  radical. 

Remove  this  factor  from  under  the  radical  sign,  extract 
the  required  root,  and  multiply  the  coefficient  by  the  root 
obtained. 


Simplify : 

1.  V28. 

2.  V72. 

3.  </n. 

4.  a/500. 

5.  -v/432. 

6.  v^l92. 

7.  a/128. 

8.  ■v/243. 

9.  \/l76. 

10.  V4'05. 

11.  2a/II2. 

12.  3^/864. 


Exercise  79. 

13.  7^/144. 

14.  S-y/mM.. 

15.  S</bW. 

16.  2\/^V. 

17.  ll^/^^^ 

18.  7\/8^. 

19.  6A/27mV. 

20.  4:^/xy. 
iJl.  a/I029. 

22.  v'-2187. 

23.  a/1250. 

24.  4\/648. 


25 


26. 


27. 


28. 


29. 


30. 


31. 


3/  64:X^y 
\27mV 

216y»" 

c,  ^\m,^n 

\243' 
^ 

\1296 
Al      512 


Sab 
2c 


228  SCHOOL   ALGEBRA. 

Case  III. 

247.  When  the  radical  expression  is  a  fraction,  the  denomi- 
nator of  which  is  not  a  perfect  power  of  the  same  degree  as  the 
radical. 

slT^^  3/     5     _   3/5  X  3  X  4_  sLq  1 

\18      \9x2      \    27x8        \  27x8 

"We  have,  therefore,  the  following  rule : 

Multiply  both  terms  of  the  fraction  by  such  a  number  as 
will  make  the  denominator  a  perfect  power  of  the  same 
degree  as  the  radical;  and  then  proceed  as  in  Case  II. 

Exercise  80. 

Simplify : 

1.  2VJ.  4.    7Vf  7.    -^1.  10.    2\/|. 

2.  8V|.  5.      ^H.  8.    VI.  11.    3^^. 

3.  iV?.  6.    3V^.  ,  9.    V^.  12.    2^A- 


14.    M.  16.    ^1-^.  18.    2<l? 

\a«  \l2bx  M3 


2a^5^c 


a;*yz' 


EADICAL   EXPRESSIONS.  229 

Case  IV. 
248.   To  reduce  a  mixed  surd  to  an  entire  surd. 

Since  aVh  =  Va"  X  ■\/b  =  -yja^b,  we  have 

(1)  3V5-V3^"^  =  V9^r5  =  V45; 

(2)  a'h^c  =  V{a'by  X  be  ^  Va*6^  xhc  =  Va*¥c ; 

(3)  2x-^xi/  =  ^/(2xy  XX])  =  a/Soj^X  xy  =  -^Sx^y  ; 

(4)  3  2/^-t^  =  a/(3  y'y  Xx'=  <fW^\ 

We  have,  therefore,  the  following  rule : 

Raise  the  coefficient  to  a  power  of  the  same  degree  as  the 
radical,  multiply  this  power  by  the  given  surd  factor,  and 
indicate  the  required  root  of  the  product. 

Exercise  81. 
Express  as  entire  surds : 

1.  5V5.           5.  2^/3.           9.  -  2-s/y.  13.  |Va. 

2.  3VIl.         6.  S</2.         10.   -3V7.  14.  -|-^a^ 

3.  S-s/S.          7.  2\/2.         11.  -m-VlO.  15.  |V^^ 

4.  2-v^.          8.  2\/4.         12.  -  2^/x.  16.  -i-y/m'. 

Case  V. 
249.   To  reduce  radicals  to  a  common  index. 
(1)    Reduce  V2  and  -^3  to  a  common  index. 

^  =  3^  =  3*=  v^2^^/9. 
Hence, 


230  SCHOOL   ALGEBRA. 

Write  the  radicals  with  fractional  exponents,  and  change 
these  fractional  exponents  to  equivalent  exponents  having  the 
least  common  denomiyiator.  Raise  each  radical  to  the  power 
denoted  hy  the  numerator,  and  indicate  the  root  denoted  by 
the  common  denominator. 

Exercise  82. 
Reduce  to  surds  of  tlie  same  order  : 

1.  Vl  and  -\/5.  7.  V2,  a/3,  and  Vb. 

2.  VU  and  V6.  8.  </^\  </b,  and  V^. 

3.  \/2  and  \/4.  9.  Va*,  ^c^  and  V^. 

4.  Va  and  V^.  10.  -y/x^y,  -Vabc,  and  -\/2^. 

5.  V5  and  V75.  11.  Va;  —  y  and  V^  +  y. 


6.    2^,  2^,  and  2\  12.    Vo+l  and  Va  -  b. 

Note.  Surds  of  different  orders  may  be  reduced  to  surds  of  the 
same  order  and  then  compared  in  respect  to  magnitude. 

Arrange  in  order  of  magnitude  : 

13.  </lb  and  V6.  15.    ^/SO,  v^,  and  VS. 

14.  -v/4  and  VS.  16.    V3,  </b,  and  </7. 

Addition  and  Subtraction  of  Radicals. 

250.  In  the  addition  of  surds,  each  surd  must  be  reduced 
to  its  simplest  form  ;  and,  if  the  resulting  surds  are  similar, 

Mnd  the  algebraic  sum  of  the  coefficients,  and  to  this  sum 
annex  the  common  surd  factor. 

If  the  resulting  surds  are  not  similar. 
Connect  them  with  their  proper  signs. 


RADICAL   EXPRESSIONS.  231 

(1)  Simplify  V27  +  V48  +  VI47. 

V27  =  (32  X  3)i  =  3  X  3^  =  3  \/3 ; 
Vi8  =  (2*  X  3)i  =  22  X  3^  =  4  X  3'  =  4V3  ; 
VT47  =  (72  X  3)i  =  7  X  3^  =  7V3. 
.'.  \/27  + Vi8  + Vl47  =  (3  +  4  +  7)\/3  =  14V3.  Ans. 

(2)  Simplify  2 a/320 -3  a/40. 

2^320  =  2(2«  X  5)i  =  2  X  22  X  5^  =  8  v^ ; 
By/¥0  =  3(23  X  5)i  =  3  X  2  X  5^  =  6  v^5. 
.-.  2  v^320  -  3  v^40  ^  (8  -  6)  v^  =  2  v^S.     Ans. 

(3)  Simplify  2VI-3VI  +  VS. 

2 V|  =  2VV-  =  2 Vi5  X  i  =  f  Vl5 ; 
3V|  =  3VJf  =  3Vl5l^^=|VT5; 

.-.  2V|-3Vf +  VS  =  (!-!  + T\)Vl5  =  iVl5.  ^ns. 

Exercise  83. 

Simplify  : 

1.  4VII  +  3Vll-5VTl. 

2.  2V3-5V3  +  9V3. 

3.  6a/4  +  2a/32  -  a/108.  7.  a/27  +  a/48  +  a/75 . 

4.  3v^  +  4a/2-a/64.  8.  4a/147  +  3V75  +  VI92. 

5.  |A/5  +  2iA/5  +  ^v^40.  9.  Va  +  iVa  +  iVa. 

6.  3a^3- 5-^48+ a/243.  10.  -y/a' -}- ^Va' ~  d</W^\ 


232  SC500L   ALGEBRA. 

11.  Va'  +  ^Va-SVa. 

12.  V256  +  2V9T  -  3 V4^. 

13.  2Vr75-3V63  +  5V28. 

14.  V2  +  3V32  +  iVl28-6VT8. 

15.  V75  + V48- VI47+ V300. 

16.  20 V245  -  V5  +  Vl25  -  2iVl80. 

17.  2V20  +  -1-VI2-2V27  +  5V45-9VI2. 

18.  7V25  +  4V45- V9-2V80+ V20-4V64, 

19.  ^5i+Vi-^250-fVf. 

20.  2V|+ V60-V15+ Vf+VS 

21.  \/277--^87+a/I25^. 

22.  V^-Vb'+VMh. 

23.  -s/aFx  +  V^  —  V4 d'h'x. 

24.  -y/^iX^yh  +  Vy*2  +  Vi*2. 

25.  Va^— aVi^  +  ^VoV. 

26.  \/8r^- •</16^+v'256^. 

27.  ^27^*- A/l25^+-^2l6^. 

28.  VSa  -  V50a' -  3 Vl8a. 

29.  6aV63^^-3Vll2^^  +  2aW343^. 


30.  3Vl25mV  +  wV20?7?- V500wV. 

31.  V32^  +  6V725  +  3Vl28a"^. 

32.  2\/^-3a^v'646  +  5a^^+2a^^^l256. 


radical  expressions.  233 

Multiplication  of  Radicals. 

251.   Since  Va  X  Vb  —  -Vab,  we  have 

(1)  8V8x5V2-3x5xV8xV2=15Vl6  =  60; 

(2)  3V2x4a/3- 3^8x4^/9  =  12^72. 
We  have,  therefore,  the  following  rule : 

Express  the  radicals  with  a  common  index.  Find  the 
product  of  the  coefficients  for  the  required  coefficient,  and  the 
product  of  the  surd  factors  for  the  required  surd  factor. 

Reduce  the  result  to  its  simplest  form. 

Exercise  84. 

1.  V3xV27.  7.   ^4x^8.  13.  V'bixV'^. 

2.  V5xV20.  8.  ■Wtx■^/'^.  14.  2V8xV2. 

3.  V2xVl8.  9.  V2xVl2.  15.  -^Sx-^^. 

4.  \^'3x-^9.  10.  VSxVe.  16.  \/7x^/^49. 

5.  \/2X\/32.  11.   -^3x^/18.  17.  \/8Tx\/=lt5. 

6.  a/27x^.  12.   -v^X-yS.  18.  |-v/l8xf-^3. 

19.  (Vl8  +  2V72-3V8)X  V2. 

20.  (\^-J-\/864  +  3\/4)x-^2. 

21.  (^V27  -  ^ V2l87  +  ^V432)  X  V3. 

22.  VSX-yi.  26.  Vfx^J.  30.  ^^xV^. 

23.  \/l6x-</250.  27.  V^  X  \/|.  31.  \/2^  X  a/^. 

24.  -{/64  X  -yi6.  28.  \/8l  X  V3.  32.  Vy''  X  V2^. 

25.  V3X\/72.  29.  ^^f  X  Vf .  33.  ^/7  X  V5. 


234  SCHOOL   ALGEBRA. 

252.    Compound  radicals  are  multiplied  as  follows : 

Ex.  Multiply  2V3  +  3V^  by  3V3-4Vi. 

2\/3  +  3\/^ 
3\/3-4\/5 

18+9V3^ 

-8\/3x-12a; 

18+    \/3^-12a; 


Multiply : 


Exercise  85. 


1.  V 5  +  V4  by  V5  -  V4.       4.  8  +  3V2  by  2  -  V2. 


2.  V9- Vl7by  V9  + VpZ.    5.  5  +  2 V3  by  3  -  5 V3. 

3.  3  +  2V5  by  2  -  V5.  6.  3  -  V6  by  6  -  3V6. 

7.  2 V6  -  3V5  by  V3  +  2V2. 

8.  7  -  V3  by  V2  4-  V5. 

9.  -v/Q- 2^4  by  4-5/3+^/2. 

10.  2V30-3V5  +  5V3by  V84-V3- V5. 

11.  3V5-2V3  +  4V7by  3V7-4V5-5V3. 

12.  4V8  +  ^Vl2-^V32by8V32-4V50-2V2. 

13.  A/6--</3+vl6by  ^/36+a/9-^4. 

14.  2V|  -  8 V|  +  3 Vf  by  3 V|  -  Vl2  -  V6. 

15.  2V|  -  4Vf  -  7Vf  by  3 V|  -  5 V30  -  2V^. 

16.  2VT2  +  8V3  +  6Viby  2VI2  +  3V3  +  6VJ. 


j  radical  expressions.  235 

Division  of  Radicals. 
253.   Since  ^  =  i%i5  =  V6,  we  have 

(1)  4^^^2V4  =  4; 
2V2 

(2)  4-</3^4v^^_4-v/3^^^_  ^/^ 
2V2      2^^         2v^« 

"We  have,  therefore,  the  following  rule : 

Express  the  radicals  with  a  common  index.  Find  the 
quotient  of  the  coefficients  for  the  required  coefficient,  and  the 
quotient  of  the  surd  factors  for  the  required  surd  factor. 

Reduce  the  result  to  its  simplest  form. 

Exercise  86. 
Divide : 

1.  V243  by  Vs.       4.  VJ  by  V|.       7.   ^^J^  by  VfJ. 

2.  \/81  by  Vs.         5.   V|  by  Vf.       8.   Vf|  by  V^- 

3.  V37'  by  V^^      6.   VS  by  Vf     9.   V||  by  Vi|. 

10.  3 V6  +  45V2  by  3 V3. 

11.  42V5  -  30V3  by  2Vl6. 

12.  84V15+168V6  by  3V21. 

13.  30V4-36Vl0  +  30V90  by  3^20. 

14.  50VI8  +  18V20-48V5  by  2V30. 

15.  V54  by  V36.     17.  VT2  by  V6.      19.  Vf  by  V3|. 

16.  V49  by  V7.       18.   V^  ^7  "V^-    20.   V2^byV«^. 
21.  V0.064  by  VIO.         22.  Var' —  y' by  a?  +  y. 


2S6  SCHOOL   ALGEBHA. 

254.  The  quotient  of  one  surd  by  another  may  be  found 
by  rationalizing  the  divisor;  that  is,  by  multiplying  the 
dividend  and  divisor  by  a  factor  which  will  free  the  divisor 
of  surds. 

255.  This  method  is  of  great  utility  when  we  wish  to 
j5nd  the  approximate  numerical  value  of  the  quotient  of 
two  simple  surds,  and  is  the  method  required  when  the 
divisor  is  a  compound  surd. 

(1)  Divide  3V8  by  V6. 

SVS     6V2     6\/2xV6     6VT2        ^  ^ 

(2)  Divide  3 V5  -  4V2  by  2V5  +  3V2. 
3V5-4\/2  _  (3\/5-4\/2)(2\/5-3\/2)  _  54-17\/l0 
2V5  +  3V2     (2\/5  +  3\/2)(2V5-3\/2)         20-18 

^54-17Vl0^27-8|VlO. 

2  ^ 

-  5 

(3)  Given  V2  =  1.41421,  find  the  value  of  -p. 

5  5V2         5\/2     7.07105 


V2      V2x  \/2        2 


=  3.53553. 


Divide:  Exercises?. 

1.  Va  +  V6  by  -yjab.  7.  3  +  5V7  by  3-5V7. 

2.  Vr25  by  5V65.,  8.  21 V3  by  4V3-8V2. 

3.  3  by  11  +  3V7.  9.  75Vl4  by  8V2  +  2V7. 

4.  3V2-lby  3V2  +  1.  10.  V5  -  VS  by  V5  +  V3. 

5.  17  by  3V7  +  2V3.  11.  V8+V7  by  V7-V2. 

6.  lbyV2-l-V3.  12.  7  — 3VIO  by  5  +  4V5. 


RADICAL   EXPRESSIONS.  237 

Given    V2  =  1.41421,    V3  =  1.73205,  V5- 2.23607; 

find  to  four  places  of  decimals  the  value  of 

13.  I^.    16.  -1-.    19.  _!-.  22.  1^=1^. 
V2       V500       3V2       5  +  4V5 

14.  -8_.    17.  -4=.    20.  -4=.  23.  5  +  V5. 

V3       V243       Vl25       V5-2 

15.  ^.         18.  -1-.     21.  ^.  24.  ^^-^. 
V5       2V3       4V5       3V2  +  1 


Involution  and  Evolution  of  Radicals.  . 

256.    Any  power  or  any  root  of  a  radical  is  easily  found 
by  using  fractional  exponents. 

(1)  Find  the  square  of  2Va. 

(2  v'a)2  =  (2  a»)2  =  22  a?  =  4  ai  =  4  v^^. 

(2)  Find  the  cube  of  2Va. 

(2  V^)3  =  (2  aif  =  23  a^  =  8  a' =  8  aVa. 

(3)  Find  the  square  root  of  4  :rV(2^6^ 

(4  xVc^^)^  =  (4  xa%^f  =  45  x^ah^  =  4^  x^a^h^  =  2  v^oTC^. 

(4)  Find  the  cube  root  of  4a;Va•^^'^ 

(4  a;Va3p)i  =  (4  a^a^i')^  =  4*  x^a^h^  =  4^  x^a^h^  =  y/Wa^^^. 

Exercise  88. 
Perform  the  operations  indicated  : 

1.  (y/mj.  3.    (Vx'y'.  5.    ^'V(:r-y)«. 

2.  (Vm')^  4.    (^)".  6.    yj-\/(a-b)''. 


238  SCHOOL   ALGEBRA. 

7.  (V2^)*.        10.   V^-  13.    ^j^/{Sa-2bf\ 

8.  (V^^^^y.    11.   ^l■^/Y2Q.  14.    \/-y32^. 

9.  {^/xy.  12.    -x/VISS.  15.    •\'l28\/243^«. 

Properties  of  Quadratic  Surds. 

257.  The  jproduci  or  quotient  of  two  dissimilar  quadratic 
surds  will  he  a  quadratic  surd.     Thus, 

Va6  X  '\/ahc=^  ah^^] 

^abc  -J-  Va^  =  "Vc. 

For  every  quadratic  surd,  when  simplified,  will  have 
under  the  radical  sign  one  or  more  factors  raised  only  to 
the  first  power ;  and  two  surds  which  are  dissimilar  cannot 
have  all  these  factors  alike. 

Hence,  their  product  or  quotient  will  have  at  least  one 
factor  raised  only  to  the  first  power,  and  will  therefore  be 
a  surd. 

258,  The  sum  or  difference  of  two  dissimilar  quadratic 
surds  cannot  be  a  rational  number,  nor  can  it  be  expressed 
as  a  single  surd. 

For  if  Va  ±  VJ  could  equal  a  rational  number  c,  we 
should  have,  by  squaring, 

azt2Vab  +  b  =  c'', 

that  is,  ±  2  Va6  =  c^  —  a  —  b. 

Now,  as  the  right  side  of  this  equation  is  rational,  the 
left  side  would  be  rational ;  but,  by  §  257,  Va6  cannot  be 
rational.     Therefore,  Va±  V^  cannot  be  rational. 

In  like  manner,  it  may  be  shown  that  Va  ±  "Vb  cannot 
be  expressed  as  a  single  surd  "Vc, 


RADICAL   EXPRESSIONS.  239 

259.  A  quadratic  surd  cannot  equal  the  sum  of  a  rational 
number  and  a  surd.  "" 

For  if  Va  could  equal  c  -\-  V6,  we  should  have,  by 
squaring, 

a  =  c*  +  2cVb  +  b, 

and,  by  transposing, 

2cVb  =  a  —  h-~c\ 

That  is,  a  surd  equal  to  a  rational  number,  which  is 
impossible. 

260.  If  a-\-  VF  =  a:  +  Vy,  then  a  mill  equal  x,  and  b 
will  equal  y. 

For,  by  transposing,  V^  —  Vy  —  x  —  a\  and  if  b  were 
not  equal  to  y,  the  difference  of  two  unequal  surds  would 
be  rational,  which  by  §  258  is  impossible. 

.*.  i  =  y,  and  a  =  x. 

In  like  manner,  if  a  —  V6  —  x—  Vy,  a  will  equal  x, 
and  h  will  equal  y. 

261.  To  extract  the  square  root  of  a  binomial  surd. 
Ex.    Extract  the  square  root  of  a  -f-  V6. 


Suppose                 Va  +  VF  =  V^  +  y/y.  (1) 

By  squaring,             a^-  y/b  =  x-\-  2Vxy  +  y.  (2) 

.-.  a  =  0?  -I-  y  and  Vft  =  2V^.  \  260 

Therefore,                 a-y/h  =  x-  2.Vxy  +  y,  (3) 


and  Va  -  V6  -  \/x  -  Vy.  (4) 

Multiplying  (1)  by  (4), 

y/a?  —  h  —  x  —  y. 
But  a^iS'\'y. 


240 


Adding,  and  dividing  by  2,  a;  =■ 
Subtracting,  and  dividing  by  2, 


SCHOOL   ALGEBRA. 


a  —  Va^  — 


+  'V 


From  these  two  values  of  x  and  y,  it  is  evident  that  this 
method  is  practicable  only  when  a^  —  6  is  a  perfect  square. 

(1)   Extract  the  square  root  of  7  +  4V3. 


Let 

v^  +  VJ  =  V7  +  4  V3. 

Then 

V^_V^  =  V7-4V3. 

Multiplying, 

x-y=  V49-48. 

.*.  a;-y  =  l. 

But 

x  +  y  =  7: 

.'.  a?  =  4,  and  y-3. 

.-.  V^  +    VJ  =  2  +  Vs. 

.  Vf+  4  V3  =  2  +  Vs. 

A  root  may  often  be  obtained  by  inspection.  For  this  purpose, 
write  the  given  expression  in  the  form  a  +  2Vb,  and  determine  what 
two  numbers  have  their  sum  equal  to  a,  and  their  product  equal  to  b. 

(2)   Find  by  inspection  the  square  root  of  75  —  12 V^I. 

It  is  necessary  that  the  coefficient  of  the  surd  be  2;  therefore, 
75  —  12  V2T  must  be  put  in  the  form 

75-2V756. 

The  two  numbers  whose  sum  is  75  and  whose  product  is  756  are 
63  and  12. 


Then 

That  is, 


75  -  2  V756  =  63  +  12  -  2V63  X  12, 

=«  ( V63  -  Vi2)». 
V63  -  V12  =  square  root  of  75  -  12  V2r; 
3  V7  -  2  V3  -  square  root  of  76  -  12  VM". 


RADICAL   EXPRESSIONS.  241 

Exercise  89. 
Find  tlie  square  root  of 

1.  7-4V3.  7.    16  +  5V7.  13.  94  +  42V5. 

2.  11  +  V72.  8.    75  +  I2V2I.  14.  11-2V30. 

3.  7  +  2VI0.  9.    19  +  8V3.  15.  47-4V33. 

4.  18  +  8V5.  10.    8V6  +  20.  16.  29  +  6 V22. 

5.  8-f2Vr5.  11.    28-16V3.  17.  83  +  12V35. 

6.  15-4VI4.        12.    51-f36V2.  18.  55  -  12 V21. 

Equations  containing  Radicals. 

262.  An  equation  containing  a  single  radical  may  be 
solved  by  arranging  the  terms  so  as  to  have  the  radical 
alone  on  one  side,  and  then  raising  both  sides  to  a  power 
corresponding  to  the  order  of  the  radical. 


Ex.  Va:'-9-f 37  =  9. 

Vo;^  -  9  =  9  -  a?. 
By  squaring,  a;'  —  9  =  81  —  18  a?  +  ic*. 

18a;  =  90. 
.*.  a;  =  5. 

263.    If  two  radicals  are   involved,  two   steps   may  be 
necessary. 

Ex.  V^  +  15-f-V^=15. 

y/x  +  15  +  V^  =  15. 
Squaring  and  simplifying,  we  have 

Va^+15a;  =  105  -  x. 
Squaring,  we  have  a;'^  + 15  a;  =  11025  -  210  a;  +  ai^. 

225  a;  =  11025. 
.\a;  =  49. 


242  SCHOOL   ALGEBRA. 


Exercise  90. 
Solve : 


1.  2V^4^  =  V28.  8.  ■v/3¥+T=3. 

2.  3V4a;-8  =  Vl3a;-3.      9.  14  + \/4a;- 40  =  10. 

3.  V^T9  =  5V^^=^.  10.   -v/l0y-4=-v/7y4-ll. 

4.  4  =  2V^-3.  11.  2^J^^=</^2{x-2)\ 

5.  5-V3y  =  4.  12.  ■y/if+~x  =  i  +  Vx. 

6.  7  +  2-y3^=5.  13.  V32+^=16-V^. 

7.  i/2x  -  3  =  -  3.  14.  V^-Va;-5  =  V6. 


15.   Vo;  +  20  -  V^  -  1  —  3  =  0. 


16.  V:??+15-7  =  7-Va7-13. 

17.  x=1~V^^^-i. 


18.    Va;  -  7  =  V:*;  +  1  -  2. 

19.  V^-3^V^+1  21.  l  +  (l-^)^^3 

V^+3      V^-2  l-(l-a;)^ 


23.  ^(a  +  ^/x-{-^la^~^/x~Vx. 

24.  Va^  —1  =  4  +  ^-y/ax  —  I-. 

25.  3V^— 3Va  =  V^— Va  +  2Va. 

26.  V9  +  2a?~  V2^  = ^ 

V9  +  2a; 


CHAPTER  XVII. 
IMAGINARY  EXPRESSIONS. 

264.  An  imaginary  expression  is  any  expression  which 
involves  the  indicated  even  root  of  a  negative  number. 

It  will  be  shown  hereafter  that  any  indicated  even  root 
of  a  negative  number  may  be  made  to  assume  a  form  which 
involves  only  an  indicated  square  root  of  a  negative  num- 
ber. In  considering  imaginary  expressions,  we  accordingly 
need  consider  only  expressions  which  involve  the  indicated 
square  roots  of  negative  numbers. 

Imaginary  expressions  are  also  called  imaginary  numbers 
and  complex  numbers.  In  distinction  from  imaginary  num- 
bers, all  other  numbers  are  called  real  numbers. 

265.  Imaginary  Square  Roots.  If  a  and  h  are  both  posi- 
tive, we  have 

I.    V^  =  Va  X  V^;         II.    ( Va)'  =  a. 

If  one  of  the  two  numbers  a  and  h  is  positive  and  the 
other  negative,  Law  I.  is  assumed  still  to  apply  ; .  we  have, 
accordingly : 

v=^ = V4(-i)  =  vfx  v^ = 2V=n;; 

V=^  =  V5(-l)  =  V5  X  V^  =  5W^ ; 

V^=  Va(-1)  =  VaX  V^^=aiV^^; 
and  so  on. 

It  appears,  then,  that  every  imaginary  square  root  can 
be  made  to  assume  the  form  aV—  1,  where  a  is  a  real 
number. 


244 


SCHOOL    ALGEBRA. 


266.  The  symbol  V—  1  is  called  the  imaginary  unit,  and 
may  be  defined  as  an  expression  the  square  of  which  is  —  1. 

Hence,     V=a  X  V=T.  =  ( V^)'  =  -  1 ; 

V—  a  X  V—  h  =  Va  X  V^^  X  V^X  V^ 

=  V^x  V^x(V-T/ 
=  V^6^x(-l) 

267.  It  will  be  useful  to  form  the  successive  powers  of 
the  imaginary  unit. 

(V^i) =  +  V^; 

(V=ly =-1; 

(V^l)»  =  ( V=l)^  V^l     =  (-  1)  V~l  =  -  V=^  ; 
( V~l)*  =  (V=l7  (V^l)'  =  (- 1)  (-  1)  =  + 1 ; 
(V^l)=^  =  ( V^l)«  v=~i     =  (+  1)  V=l  =  +  V^l ; 
and  so  on.     We  have,  therefore, 

(V^ir+^=+v=T; 

268.  Every  imaginary  expression  may  be  made  to  assume 
the  form  a  +  ^V—  1,  where  a  and  h  are  real  numbers,  and 
may  be  integers,  fractions,  or  surds. 

If  Z>  =  0,  the  expression  consists  of  only  the  real  part  a, 
and  is  therefore  real. 

If  a  =  0,  the  expression  consists  of  only  the  imaginary 
part  £V—  1,  and  is  called  a  pure  imaginary. 


IMAGINARY   EXPRESSIONS.  245 

269.  The  form  a  +  6  V—  1  is  the  typical  form  of  imaginary- 
expressions. 

Reduce  to  the  typical  form  6  +  V— 8. 

This  maybe  written  6+V8xV^,  or  6  +  2V2xV^; 
here  a  =  6,  and  h  —  2 V2. 

270.  Two  expressions  of  the  form  a-|-5V— 1,  a—hsf—Y^ 
are  called  conjugate  imaginaries. 

To  find  the  sum  and  product  of  two  conjugate  imagi- 
naries, 

a-5V^ 


The  sum  is 

2a 

a  +  ^>  V-  1 

a  -  5  V-  1 

, 

a'  +  aW-l 

-aW-1  +  5* 

The  product  is 

a^                    +5' 

From  the  above  it  appears  that  the  sum  and  'product  of 
two  conjugate  imaginaries  are  both  real. 

271.   An  imaginary  expression  cannot  he  eqvxil  to  a  real 
number. 

For,  if  possible,  let 

a  +  5  V^  =  c. 
Then  transposing  a,  6  V—  1  =  c  —  a, 
and  squaring,  —b^=^(c  —  a)'.       ^ 

Since  b'^  and  (c  —  af  are  both  positive,  we  have  a  nega- 
tive number  equal  to  a  positive  number,  which  is  impossible. 


246  SCHOOL    ALGEBRA. 

272.  If  two  imaginary  expressions  are  equal,  the  real  parts 
are  equal  and  the  imaginary  parts  are  equal. 

For,  let  a  +  h^^l  =  c  +  c?V^. 

Then  (h- d)^-l==  c-a\ 

squaring,  —(b  —  dy  =  (c  —  a)', 

which  is  impossible  unless  b  =  d  and  a  =  c. 

273.  Ifx  and  y  are  real  and  a:-f-yV— 1=0,  then  x=0 
and  y  =  0. 

For,  yV—  1  =  —  x, 

-y'  =  x\ 

^^  +  y'  =  o, 

which  is  true  only  when  a;  =  0  and  y  =  0. 

274.  Operations  with  Imaginaries. 

(1)  Add5  +  7V^and8-9V=^. 

The  sum  is  5  +  8  +  7  V^  -  9\/^, 

or  13  -  2v^ri. 

(2)  Multiply  3  +  2  V^  by  5  -  4  V^. 

(3  +  2\/:rT)(5-4\/^) 

=  15  -  12\/^  +  10\/=1  -  8(-  1) 
=  23-2v/I^. 

(3)  Divide  14  +  5 V=^l  by  2  -  3 V^. 

14  4  5  Vn     (14  +  5  V^X)  (2  +  3  V^) 
2_3V=1      (2-3V^)(2  +  3V^) 

13  +  52v^ 
4-(-9) 

_13  +  52V3 
'    "  13 

=  l+4\/^. 


IMAGINARY  EXPRESSIONS. 


247 


Exercise  91. 
Reduce  to  the  form  5  V^  : 


1.    V-9. 


9.    V-625. 


2.  v=n^. 

3.  V^^25. 

4.  V=ni4. 

5.  V=^^l69. 

6.  V-a:*. 

7.  v=^8r. 

8.  V^=^256. 
Add: 


10.  V-36. 

11.  ^^. 

12.  ■v/^=^729. 


13.    V-  289. 

14.  '-s/^^im. 


15.  V=^^. 

16.  V^^'. 


17.  </~x^. 

18.  V^. 

19.  V-  a'b-\ 

20.  V-  9a;*. 

21.  \/-(2a;-3yy^. 

22.  V-  (a:  -  2y)^. 

23.  v^^c^+y). 


24.    W—(x^  —  f). 
25.   V=^  +  V^=^  -  V^=^T2i. 


26.    V-  64  +  V-  1  -  V-  36. 
27. 


"^^^  +  V- 4  a*  +  V- 16a* 


28.  V-a'+ V-81a'-V-a'. 

29.  a-iV^  +  a  +  5V^. 
-     30.    2  +  3V=n-2  +  3V^. 

31.  a  +  5V^  +  c  — c?V^. 

32.  3aV^-(2a  — 6)V^. 
Multiply : 

33.  V^  by  V^.  36.    V^  by  V^. 

34.  -  V^  by  V^=^.  37.    V^^  by  V-^. 

35.  V^T6  by  V^.  38.    V^  by  V=T6. 


248 


SCHOOL   ALGEBRA. 


39. 
40. 


V^=^  by  V^^^^.  41.    3V^  by  2V^. 

■V-ia  +  b)  by  V-(a-b).    42.    -5V^  by  2V^. 
43.    V^^  +  V^  by  V^^  -  V^. 


44.    X 


1  +  V  -  3  ,  1 

—^ by  X 


2 

45.  a^—a-\-b^/—b  by  a'V—a  —  b^—b. 

46.  2V^  +  3V=^  by  3V=^-2V^. 

47.  V3  +  2V=^  by  V3-2V^. 

48.  m  — 3V^^  by  w  +  4V^. 


Perforin  the  divisions  indicated : 


49. 


50. 


51. 


52. 


53. 


54. 


v= 

T 

b 

v= 

rp 

c 

v= 

4 

v: 

-9 

v= 

81 

v= 

■a 

v= 

^ 

V- 

-aa; 

v: 

-a; 

65. 


56. 


57. 


58. 


59. 


60. 


-V—x 

X 

V-  10  rr' 


61. 


62. 


63. 


64. 


65. 


1 


3_V^ 
2  +  V^ 


a-\-x- 


V—  5  a; 
2V?  ' 


a  —  a;  V—  1 

V5  + V^ 

2a  +  35V^" 
2a-35V^ 


66.   \—^^-J-^ 
^a-\bV^^ 


/y 


CHAPTER  XVIII. 
QUADRATIC    EQUATIONS. 

275.  "We  have  already  considered  equations  of  the  first 
degree  in  one  or  more  unknowns.  We  pass  now  to  the 
treatment  of  equations  containing  one  or  more  unknowns 
to  a  degree  not  exceeding  the  second.  An  equation  which 
contains  the  square  of  the  unknown,  but  no  higher  power, 
is  called  a  quadratic  equation. 

276.  A  quadratic  equation  which  involves  but  one  un- 
known number  can  contain  only  : 

(1)  Terms  involving  the  square  of  the  unknown  number. 

(2)  Terms  involving  the  first  power  of  the  unknoM^n 
number. 

(3)  Terms  which  do  not  involve  the  unknown  number. 

Collecting  similar  terms,  every  quadratic  equation  can  be 
made  to  assume  the  form 

where  a,  h,  and  c  are  known  numbers,  and  x  the  unknown 
number. 

If  a,  5,  c  are  numbers  expressed  by  figures,  the  equation 
is  a  numerical  quadratic.  If  a,  h,  c  are  numbers  represented 
wholly  or  in  part  by  letters,  the  equation  is  a  literal  quadratic. 

277.  In  the  equation  ao:^-{-hx-{-c^=0,  a,  5,  and  c  are 
called  the  coefficients  of  the  equation.  The  third  term  c  is 
called  the  constant  term. 


250  SCHOOL   ALGEBRA. 

If  the  first  power  of  x  is  wanting,  the  equation  is  a  pure 
quadratic ;  in  this  case  6  =  0. 

If  the  first  power  of  x  is  present,  the  equation  is  an 
affected  or  complete  quadratic. 

Pure  Quadratic  Equations. 
278.   Examples. 

(1)  Solve  the  equation  5a;'  —  48  =  2x\ 
We  have  5  a;^  -  48  =-.  2  a;2. 
Collect  the  terms,  3  x^  =  48.  . 
Divide  by  3,                               a;^  =  16. 
Extract  the  square  root,            a;  =  ±  4. 

It  will  be  observed  that  there  are  two  roots,  and  that  these  are 
numerically  equal,  but  of  opposite  signs.  There  can  be  only  two 
roots,  since  any  number  has  only  two  square  roots. 

It  may  seem  as  though  we  ought  to  write  the  sign  ±  before  the  x 
as  well  as  before  the  4.  If  we  do  this,  we  have  +a;=  +  4,  — »=»  —  4, 
+  x  =  —  4,  — a;=  +  4. 

From  the  first  and  second  equations,  a;  =  4 ;  from  the  third  and 
fourth,  a;  =  —  4;  these  values  of  x  are  both  given  by  the  equation 
a;  =  ±  4.  Hence  it  is  unnecessary  to  write  the  ±  sign  on  both  sides  of 
the  reduced  equation. 

(2)  Solve  the  equation  3a:'  —  15  =  0. 
We  have  3a;2=15, 

or  a;2  =  5. 

Extract  the  square  root,  a;  =  ±  VE. 

The  roots  cannot  be  found  exactly,  since  the  square  root  of  5  can- 
not be  found  exactly ;  it  can,  however,  be  determined  approximately 
to  any  required  degree  of  accuracy;  for  example,  the  roots  lie  between 
2.23606  and  2.23607  ;  and  between  —  2.23606  and  —  2.23607. 

(3)  Solve  the  equation  Sic^-j-  15=0. 
We  have  Sx^=  —  15, 

or  x^=  —  5. 

Extract  the  square  root,  a;  =  *  -v^  — 6. 


QUADRATIC   EQUATIONS.  251 

There  is  no  square  root  of  a  negative  number,  since  the  square  of 
any  number,  positive  or  negative,  is  necessarily  positive. 

The  square  root  of  —5  differs  from  the  square  root  of  +  5  iti  that 
the  latter  can  be  found  as  accurately  as  we  please,  while  the  former 
cannot  be  found  at  all. 

279.  A  root  which  can  be  found  exactly  is  called  an  exact 
or  rational  root.  Such  roots  are  either  whole  numbers  or 
fractions. 

A  root  which  is  indicated  but  can  be  found  only  approx- 
imately is  called  a  surd.  Such  roots  involve  the  roots  of 
imperfect  powers. 

Rational  and  surd  roots  are  together  called  real  roots. 

A  root  which  is  indicated  but  cannot  be  found,  either 
exactly  or  approximately,  is  called  an  imaginary  root.  Such 
roots  involve  the  even  roots  of  negative  numbers. 


Exercise  92. 
Solve : 

1.  Sx'  —  2  =  x'-{-Q.  Q     S-x'.x'  +  b     , 

~n — '     A — ~^ 

2.  5a;' +10 -6:1:^+1.  ^^  ^ 

3.  7x'-50-=4:x'  +  25.  10.    5^_±_3_17-^' 


4.     Qx'-^  =  4:X'  +  -^-^- 


■g-  — t:^  -r-g-- 


8 

^2  _4_  1  11      _§ 1_  _  1. 

^-^-=10.  ^^'    ^a^      6x^-S 

3^'-8      ,  12.    A_    3   __i 


e.  :-^^  =  4. 


3a;'     bx"      15 


x'-9_^x'-hl  '         13.    _1 L_  =  l. 

4  5     *  *   a;— 1     a;+ 1      4 

7^6  8-a:^2-3» 


252  SCHOOL   ALGEBRA. 

15.  Sx'  +  Ux^lOx  +  S  +  x'  +  x. 

16.  (x  +  A)(x  +  5)  =  S(x+l)(x  +  2)-^. 

17.  S(x~2)(x  +  S)^(x+l)(x  +  2)  +  x'  +  5. 

18.  (2x  +  1)  (3a;  -  2)  +  (1  -  a;)  (3  +  4:x)  =  Sx''-  15. 

^^-   "17  9~  +  "~5""-^^- 

20.    _^_  +  _^  2      -5. 

lQa;^  +  7      12a:^  +  2_^5:r^-9 
18  lla;^-8  9 

22     ^-1  .  ^  +  1_5  26.    -  + 


X      a  ax 


x+l      x-l      2 

23.    ax'  +  b^c.  27.    ^+^  +  ^::i^  =^ 5. 

.-z;  —  a     X  -{-  a      2 

28       25        5a;  +  2^>_      ^ 
25.    a;'-[-26a:+c  =  a>(2a7  +  l).  *    :r-6  3a7 

29.  2{(x+a)(x'}-b)  +  {x-a')(x-b)l^a'  +  U\ 

30.  25(ar  -  a)  (or  4-  ^)  +  (^  +  a)  (^  -  ^)S  =  9a'^  +  2aZ»  +  ^>^ 

Affected  Quadratic  Equations. 

280.  Since  (x  ±:  by  =  x^  ±2bx +  b\  it  is  evident  that 
the  expression  x^±2bx  lacks  only  the  third  term,  b^,  of 
being  a  perfect  square. 

This  third  term  is  the  square  of  half  the  coefficient  of  x. 

Every  affected  quadratic  may  be  made  to  assume  the  form 
x'  dz2bx  =  c,hy  dividing  the  equation  through  by  the  co- 
efficient of  a;'. 


QUADRATIC   EQUATIONS.  253 

To  solve  such  an  equation : 

The  first  step  is  to  add  to  both  members  ike  square  of 
half  the  coefficient  ofx.    This  is  called  completing  the  square. 

The  second  step  is  to  extract  the  square  root  of  each  mem- 
ber of  the  resulting  equation. 

The  third  step  is  to  reduce  the  two  resulting  simple 
equations. 

(1)    Solve  the  equation  x^  —  ^x  =  20. 

We  have  a;2-8a;  =  20. 

Complete  the  square,  a;''— 8a;  +  16  =  36. 
Extract  the  square  root,  a;  —  4  =  ±  6. 
Reduce,  ic  =  4  +  6  =  10, 

or  a;  =  4-6  =  -2. 

The  roots  are  10  and  — -  2. 

Verify  by  putting  these  numbers  for  x  in  the  given  equation : 


a;  =10, 

102 -8  (10)  =  20, 

100-  80  =  20. 


a;  =  -2, 
(-2)2 -8  (-2)  =  20, 
4  +  16  =  20. 


(2)   Solve  the  equation  ^"^    =  — — — • 

X  —  L      X  ~j~  y 

Free  from  fractions,  {x  +  1)  (a;  +  9)  =  (a;  —  1)  (4  a;  —  3). 
Simplify,  3a;2-17a;  =  6. 

We  can  reduce  the  equation  to  the  form  a;'  — 26a;  by  dividing  by  3. 
Divide  by  3,  a;2_Y-x  =  2. 

Half  the  coefficient  of  a;  is  ^  of  —  ^  =  —  -y-»  ^°^  ^^  square  of  —  V 
is  ^^'.     Add  the  square  of  —  Y-  to  both  sides,  and  we  have 

3        U  J  36 


17  19 

Extract  the  root,  a;  -  -^  =■  ±  — • 

D  D 


254 
Reduce, 


SCHOOL    ALGEBRA. 


.-1-7  =  , 19. 
6  6 


17     19 


The  roots  are  6  and 

3 

Verify  by  putting  these  numbers  for  x  in  the  original  equation 


a;  =  6. 
6  +  1  _  24  -  3 
6-1       6  +  9' 

7^21 

5     15 


3 

-Ul      -4_3 
3  3 


i-1 


3 
13 

26* 


+  9 


Solve : 
1.  x'+2x  =  S. 

3.  a;' -4a;  =  12. 

4.  X^-{-4:X=  5. 


Exercise  93. 

5.  ^^  +  5a;=14. 

6.  x''-2>x  =  2%. 

7.  2a;'  +  ;r=15. 

8.  ^x^-]-2>x  =  2. 


9.  a;2  +  |a;=40. 

10.  3:?;' -4^7  =  4. 

11.  6a:' +  ^=1. 

12.  ^x^-x  =  2. 


13.  12a;'^— lla:  +  2  =  0. 

14.  15a;''- 2a:- 1=0. 

15     (:r+l)(:r  +  2)      ix-l)(x-2)_^ 
5  2 

16.    (2:r-3)a:      (a;  + 4)(a:- 1)_  ^ 
4  6 


QUADRATIC  EQUATIONS.  255 

^^     3a?  +  5  ■  2a:  — 5^g  22     x-\-2     4  — a;_7 


18. 


x  +  4:        x-2  x-\        2x       3 

X—  6  ■    37  +  5  _  1  20     £+_3_5^_+8 

a;-2      2a;+l~    *  '   a;-2'~a;  +  4" 


19  4-3a7      l  +  2a7_9  2a7-l  .        3      __ 

2  +  x        l-x  ~  i  *        8      '^ 2x-\ 

20  -^-  +  ^±i  =  l^.  25  ^  +  1  I  ^  +  2_13 
x^\         X         6  *  x-\-2     x-\-\      6 

21.    bx^  —  ^x  =  \.  26.  7a;2_3^^_]__ 


Another  Method  of  Completing  the  Square. 

281.  When  the  coefficient  of  a?  is  not  unity,  we  may- 
proceed  as  in  the  preceding  section,  or  we  may  complete 
the  square  by  another  method. 

Since  (ax  ±  ISf  =  aV  =t  2  abx  +  Z>^  it  is  evident  that  the 
expression  a^x^±  2ahx  lacks  only  the  third  term,  Z>',  of  being 
a  complete  square. 

It  will  be  seen  that  this  third  term  is  the  square  of  the 
quotient  obtained  from  dividing  the  second  term  hy  twice  the 
square  root  of  the  first  term. 

282.  Every  affected  quadratic  may  be  made  to  assume 
the  form  of  aV  zb  2  abx  =  c. 

To  solve  such  an  equation : 

The  first  step  is  to  complete  the  square;  that  is,  to  add  to 
each  side  the  square  of  the  quotient  obtained  from  dividing 
the  second  term  by  twice  the  square  root  of  the  first  term. 

The  second  step  is  to  extract  the  square  root  of  each  side 
of  the  resulting  equation. 

The  third  and  last  step  is  to  reduce  the  two  resulting 
simple  equations. 


256  SCHOOL   ALGEBRA. 

283.  Examples. 

(1)  Solve  tlie  equation  16x^-\-5x  —  S  =  7 x^  —  x  +  ib. 

"We  have  IQx^ -\- 5x— 3=7  x^  —  x-h  ^5. 

Simplify,  9a;2+6a;=48. 

The  square  root  of  dx^  is  3  a;,  and  twice  3  a;  is  6  a;,  The  second 
term  divided  by  6  a;  is  1.     Square  1  and  add  it  to  both  sides. 

9a;2  + 6a; +1  =  49. 
Extract  the  square  root,  3  a;  + 1  =  ±  7. 
Reduce,  3a;=— l  +  7or— 1  —  7, 

.'.x=  2  or  —  2f. 

Verify  by  substituting  2  for  x  in  the  equation : 

16a;2  +  5a;  -  3  =  7a;2  -  a;  +  45, 

16  (2)2  +  5  (2)-  3  =  7  (2)2  -  (2)  +  45, 

64+10-3  =  28-2  +  45, 

71  =  71. 

Verify  by  substituting  —  2f  for  x  in  the  equation: 
16a;2  +  5  a;  -  3  =  7a;2  -  a;  +  45, 

K-i;-(-i)— v(-i)^-(-i)-^. 

1024      40      ^      448  ,  8  ,   ,, 

"9 ¥-^  =  —+3  +  ^^' 

1024  -  120  -  27  =  448  +  24  +  405, 

877  =  877. 

(2)  Solve  the  equation  3:r^  —  4:r  =  32. 

Since  the  exact  root  of  3,  the  coefficient  of  a;2,  cannot  be  found,  it 
is  necessary  to  multiply  or  divide  each  term  of  the  equation  by  3  to 
make  the  coefficient  of  x^  a  square  number. 

Multiply  by  3,  9  a;2  —  12  a;  =  96. 

Complete  the  square, 

9a;2-l2a;+4=100. 
Extract  the  square  root,  3  a;  —  2  =  ±10. 
Reduce,  3a;=  2  +  10  or  2— 10; 

.-.  a;  =  4or-21. 


QUADRATIC   EQUATIONS.  257 


Or,  divide  by  3,  a^-i^  =  — • 

•^     '  3        3 

Complete  the  square,  a;2-iH  +  i  =  ^  +  i  =  M 

0*7     0     17      y 

2        10 

Extract  the  square  root,     x =  ±  — 

o  o 

2±10 


=  4  or  -  2|. 
Verify  by  substituting  4  for  x  in  the  original  equation : 
48  -  16  =  32, 
32  =  32. 

Verify  by  substituting  —  2f  for  x  in  the  original  equation ; 
21i-(-10f)  =  32, 
32  =  32. 

(3)   Solve  the  equation  —Sx'^-{-5x  =  —  2. 

Since  the  even  root  of  a  negative  number  is  impossible,  it  is  neces- 
sary to  change  the  sign  of  each  term.    The  resulting  equation  is 

3x2 -5a;  =  2. 


Multiply  by  3, 

9x2 -15a;  =  6. 

Complete  the  square, 
9x2- 

-15x  +  25^49. 
4       4 

Extract  the  square  root,  3  a;  —  -  =  ±  ^• 
2        2 

Reduce, 

3.  =  Sf. 

3a;  =  6  or -1. 

...a;^2or-|. 

Or,  divide  by  3, 

•t     5x     2 
3       3' 

Complete  the  square, 
a:» 

5«;     25  _  49 
3       36  ^  36 

258  SCHOOL   ALGEBRA. 

Extract  the  square  root,     x 


5 
6 

=^|- 

X 

5±7 
^    6   ' 

=  2or- 

1 
■3* 

284.  If  the  equation  Sx^—  5a?  =  2  be  multiplied  hy  four 
times  the  coefficient  of  x^,  fractions  will  be  avoided. 

We  have  36  a;^  -  60  a;  =  24. 

Complete  the  square, 

36  a;2- 60  a; +  25  =  49. 
Extract  the  square  root,  6a;  —  5  =  ±  7. 

6a;  =  5±7. 

60;  =  12  or -2. 

.-.  a?=2or-i- 
3 

It  will  be  observed  that  the  number  added  to  complete  the  square 
by  this  last  method  is  ihe,  square  of  the  coefficient  of  x  in  the  original 
equation  3  a;^  —  5  a;  =  2. 

Note.  If  the  coefficient  of  x  is  an  even  number,  we  may  multiply 
by  the  coefficient  of  x^,  and  add  to  each  member  the  square  of  half  the 
coefficient  of  x  in  the  given  equation. 

(1)   Solve  the  equation  4  a;'  —  23  a;  =  -  30. 

Multiply  by  four  times  the  coefficient  of  x^,  and  add  to  each  side 
the  square  of  the  coefficient  of  x,      ^ 

64a;2  -  ( )  +  (23)2  _  529  _  430  =  49. 
Extract  the  square  root,  8  a;  —  23  =  ±  7. 
Reduce,  8a;  =  23±7; 

8a?  =  30  or  16. 
.-.  a;  =  3f  or  2. 

Note.  If  a  trinomial  is  a  perfect  square,  its  root  is  found  by  taking 
the  roots  of  the  first  and  third  terms  and  connecting  them  by  the  sign 
of  the  middle  term.  It  is  not  necessary,  therefore,  in  completing  the 
square,  to  write  the  middle  term,  but  its  place  may  be  indicated  as 
in  this  example. 


QUADRATIC   EQUATIONS.  269 

(2)   Solve  the  equation  72  a;'  —  30  a;  =  —  7, 

Since  72  =  2'  X  3',  if  the  equation  is  multiplied  by  2,  the  coeffi- 
cient of  a;'  in  the  resulting  equation,  144a;2  — 60a;  =  — 14,  will  be  a 
square  number,  and  the  term  required  to  complete  the  square  will  be 

I  —  )  ={-]  =  — .     Hence,  if  the  original  equation  is  multiplied  by 

4x2,  the  coefficient  of  x^  in  the  result  will  be  a  square  number,  and 
fractions  will  be  avoided  in  the  work.     We  shall  then  have 

576x2 --240  a;  =  -56. 
.-.  576a;2-()4-25  =  -31. 
Extract  the  root,  24  a;  -  5  =  ±  V^^. 

a;-^V(5±V=^). 


Solve:  Exercise  94. 

1.   2>x^-2x  =  ^.  ^^    30:^  +  ^  =  25. 


2.  5a;'-6a7  =  27. 

3.  2s^-\-Zx  =  b. 

4.  2a;' -6a;  =  7. 

5.  3a;'+7a;=6. 

6.  6a;*^-7a;  =  24. 

7.  8a;' +  3a;  =  26. 

8.  7a;' +  6a;  =150. 

9.  6a;'  +  5a;  =  14. 

10.  7a;'-2a;  =  f. 

11.  8a;'+7a;  =  61. 

12.  7a;' -20a;  =76. 

13.  11a;'- 10a;  =  24. 


3 


15. 

^2     3^-3^+1. 
4 

16. 

f-|-2(.-2). 

17. 

¥+t-- 

18. 

¥-=-  =  ^ 

19. 

^   2  ,   ^        20 
1^  M-f^^y 

20. 

2a;-3=?. 

X 

21. 

Ix      6  _20 
5       3a;      3* 

22. 

2a;       3       10 
3"^2a;      3* 

260  SCHOOL   ALGEBRA. 

23.  (:r  +  2)(2a;+l)  +  (a;-l)(3:r  +  2)  =  57. 

24.  Sx(2x  +  b)~(x  +  S)(Sx-l)^l. 

(2x-}-5)(x-S)     x(Sx  +  ^) 

3  "^         5  • 

26.    i(5x'-8x-6')-i(x'  —  S)  =  2x  +  l. 
27.    -?-  +  5=2.  30.    -?_=:_1_+     2 


^  +  3     a;  x—1      x  —  2     x  —  A 

28.  -^ ^  =  i  31.  ^+2  +  ^±3^_5. 

a;-l      2a;+l      3  a;-4      :r-2 

29.  -J_  +  -4       ^5         3^^  2^-3      5-^^ 
3a;-2      2a;-5  a;-4        a;  +  2 

33     ll-3rg  ■  2(7-4:r)_.. 
1-x  l-2a; 

3^     a;+l    ,  l--x_        2 


:r^-4     a;4-2      5(a;-2) 
35.    2^  +  I  +  i^:^^5. 

36        237  +  3  7-37    __7-3a; 

2(237-1)      2(37  +  1)      4-337' 

37.  2^=i L:-=^Zi2  ,  5 

3  37  —  8        37-8 

38.  337  +  2  7-37^737-1 
237-1^23:+l        437^-1^      ■ 


37-5    ,37-8  80 


37+3        37-3        37^-9        2 

40      237+1       437+l_       45  . 

'       7-37"^    7  +  37         49-37='"^     • 


quadratic  equations.  261 

Literal  Quadratics. 
285.  Examples. 

(1)  Solve  the  equation  ax' -{- bx -\- c  =  0. 

Transpose  c,  ax^  +  bx  =  —  c. 

Multiply  the  equation  by  4  a  and  add  the  square  of  b, 

4aV  +  ()  +  62  =  62_4ac. 
Extract  the  root,  2ax  +  b  =  ±  Vb^  —  4ac. 

.    ^      —b±  y/b"^  —  4  ac 

. .  X  = ' — 

2a 

(2)  Solve  the  equation  (a'  -{- 1)  x  =  ax' -{•  a. 
Transpose  ax^  and  change  the  sign, 

ax"^  ~{a^ +  l)x  =  —  a. 
Multiply  by  4  a,  and  complete  the  square, 

4aV-(  )  +  (a2  +  1)2  =  _  4a2  +  a*  +  2a2  +  1 
=  a*-2a2  +  i. 

Extract  the  root,    2aa;  —  (a'  + 1)  =  ±  {a^  _  l). 
Reduce,  2ax^  (a^  +  1)  ±  (a«  -  1), 

=  2a2or2. 

.'.  x  =  a  or  — 
a 

(3)  Solve  the  equation  adx  —  acx^  —  bcx  —  hd. 

Transpose  bcx  and  change  the  signs, 

acx^  +  bcx  —  adx  =  bd. 

Express  the  left  member  in  two  terms, 

acx^  +  {bc  —  ad)x='  bd. 

Multiply  by  4  ac,  and  complete  the  square, 

4a2cV  +  ( )  +  (&c  -  acf)2  =  JV  +  2  abed  +  aH\ 

Extract  the  root, 

2acx  +  {be  —  ad)  =  ±  (6c  +  ad). 

Reduce,  2 acx  =-  —  {bc  —  ad)±  (be  +  ad) 

—  2ad  or  —  26c. 

d  b 

,'.  a;  —  -  or  —  -. 

c  a 


262  SCHOOL  ALGEBRA. 


(4)   Solve  the  equation  px^  —  px-\-  qa^  -\-  qx  — 
sft  member  in  two  ter\ 
(p  +  q)x^-{p-q)x 


Express  the  left  member  in  two  terms, 

pq 


p  +  q 
Multiply  by  four  times  the  coefficient  of  x\ 

4(j3  +  qfx'  —  4(p2  _  q'i)x  =  4ipq. 
Complete  the  square, 

Hp  +  qy^-i)  +  (i>  -  qy==p^  +  2pq  +  q*- 

Extract  the  root, 

2{p  +  q)x-{p-q)=^±(p  +  q). 
Reduce,  ^{p  +  q)^  =  ip  —  2)  ^  {p  +  q) 

=  2p  or  —  2  2'. 

or 


p  +  g-  p  +  q 

Note.  The  left-hand  member  of  the  equation  when  simplified 
must  be  expressed  in  two  terms,  simple  or  compound,  one  term  con- 
taining x^,  and  the  other  term  containing  x. 

Exercise  95. 

Solve : 

1.  x'-{-2ax  =  Sa\  9.  2aV  +  a2r— 1  =  0. 

2.  .7^-4:ax  =  12a\  ^O.  12b^x^- bbx=S. 

3.  x'  +  8bx  =  9b\  11.  ^  +  ^=na(x-Sa). 

4.  x'-{-Sbx  =  10b\  o   2  o 

12.  i£V-^  =  A. 

5.  x'  +  5ax  =  Ua\  4        2a     2a' 

6.  Sx'  +  Acx  =  4:c\  13.   x'--  =  ~ 

a      4  a' 

7.  5ax-2x'  =  2a\  o     ,      ^ 

14     3aa;'  .  2a? ^13 

8.  6x'-ax-a'  =  0.  '      4         3       3a* 

-_     x'-i-ax-^a^  ,  a'—x* 


QUADRATIC   EQUATIONS.  263 

16.    2a;'-a;  +  a  =  2a'.  22.  x^  +  (a-h)x  =  ah. 

^^     x{a-x)  J  s;^^^  23.  ^^-il£  +  ^LZL^  =  §. 

a-{-x        3          *  '  2a  — X     a-\~2x     3 

^g     a:(3a;-a)^  a  2^  2£-3a  ,  3£-f2a ^  10^ 

^x-\-a         12  *  x  +  ^a        4a:  — a        7 

^  ,  m^  —  'r?        -  „„  a;'        5a;    ,    1      ^ 


^nn 


65^      6a6      a'^ 


20.    ^±«  +  ^_II«  =  2.  26.    ^-^  +  ^+±±i  =  2. 

^  —  a     X -\- a  a-\-b-\-x      x-\-b 

36  —  X     x-j-Za  a-f-o-{-x      a      b      x 

gg     {2>x-a){2x-\-a)  .  a'-4a:^^o^a 
2  3 

29.  9a;^  — 3(a  +  25)a;+2a5  =  0. 

30.  (2a+l)a;^  +  3a2a;  +  a'  — a'  =  0. 

31.  {l-a')x''-2(l  +  a')x-{-l-a^  =  0, 

32.  (a  +  ^>)V-  (a^  -  Z>'0 a;  =  ab. 

33.  aV  +  25ca;+a'  =  cV4-2a'a;+5^ 

34.  (a  +  Z>)a;'  — (2a  +  Z»)a;  +  a  =  0. 
2a;  — 35         3a;  a 


35 


36. 


x-2b      x  +  2a   ~2{a-b) 
Sx-2b      4a;-|-25      a-b 


X  —  h  x-^-a        a-{-b 

37.  (3a^4-5^)(a;'-a;+l)  =  (a'  +  35^)(a;'  +  a:  +  l). 

38.  x'—{a-\-b)x-\-ac-\-bc  —  c''  =  0. 

39.  x^  —px  =  {p-^q-^r){q  +  r). 

40.  (a'  +  a5)(a;»-l)-(a'  +  i')a;  +  (a  +  J)(a-25)  =  a 


264  school  algebra. 

Solutions  by  Factoring. 

286.  A  quadratic  which  has  been  reduced  to  its  simplest 
form,  and  has  all  its  terms  written  on  one  side,  may  often 
have  that  side  resolved  hy  inspection  into  factors. 

In  this  case,  the  roots  are  seen  at  once  without  com- 
pleting the  square. 

(1)  Solve  a^-\-1x-m  =  (). 

Since  a;^  +  7a:  -  60  =  (a;  +  12) (a;  -  5), 

the  equation  ,       a;''  +  7  a;  —  60  =  0 

may  be  written     '  {x  +  12)  (a;  —  5)  =  0. 

If  either  of  the  factors  a;  +  12  or  a;  —  5  is  0,  ih.Q  product  of  the  two 
factors  is  0,  and  the  equation  is  satisfied. 

Hence,  a;  +  12  =  0,   or    a;  —  5  =  0. 

.*.  a;  =  —  12,    or    a;  =  5. 

(2)  Solve  2a?-x'  —  ^x  =  0. 

The  equation  2a;^  —  a;'^  —  6a;  =  0 

becomes  a;  (2  a;^  —  a;  —  6)  =  0, 

and  is  satisfied  if      a;  =  0,  or  if  2  x'  —  a;  —  6  =  0. 

3 

By  solving  2^2  —  a;  —  6  ^  0,  the  two  roots  2  and are  found. 

3  ^ 

Hence  the  equation  has  three  roots,  0,  2, 

(3)  Solve  3e'-\-x^~A:X-^  =  0. 

The  equation  a^^+a^^  — 4a?  — 4=.0 

becomes  a^  (a;  +  1)  —  4  (a;  +  1)  =  0, 

(a;2-4)(a;  +  l)  =  0. 
Hence  the  roots  of  the  equation  are  —  1,  2,  —  2. 

(4)  Solve  ^'  -  2^:^  -  11  a;  +  12  -  0. 

By  trial  we  find  that  1  satisfies  the  equation,  and  is  therefore  a 
root  (^  89). 

Divide  by  a;  —  1 ;  the  given  equation  may  be  written 
{x-l){x^-x -12)^0, 
and  is  satisfied  if  a;  —  1  =  0,  or  if  a;*  —  a;  ^  12  =»  0. 

The  roots  are  found  to  be  1,  4,  --  3, 


(QUADRATIC   EQUATIONS.  265 

(5)   Solve  the  equation  x{x'^  —  9)  =  a(a^  —  9). 

If  we  put  a  for  x,  the  equation  is  satisfied ;   therefore  a  is  a 
root  il  89). 

Transpose  all  the  terms  to  the  left-hand  member  and  divide  by 
x  —  a. 

The  given  equation  may  be  written 

{x  -  a){x^  +  oa;  +  a'  -  9)  =  0, 

and  is  satisfied  if  a;  —  a  =■  0, 

or  if  a;2  +  arc  +  a^  -  9  =  0. 

The  roots  are  found  to  be 


-a-h  V36-3a2     -a-  \/36-3o« 
^'  2 '    2 


Exercise  96. 
Find  all  the  roots  of 

1.  a;'-5a;  +  4  =  0.  5.   a;' +  ar^- 6a?  =  0. 

2.  6a;''-5a;-6  =  0.  6.    0^-^  =  0. 

3.  2a;'- ar- 3  =  0.  7.   a;'  +  8=-0. 

4.  10a;' -fa;- 3  =  0.  8.    a;*-16  =  0. 

9.    (a;-l)(a;-3)(a;'  +  5a;-f  6)  =  0. 

10.  (2a;  -  1) (a; -2)  (3a;' -5a;- 2)  =  0. 

11.  (a;'  +  a;-2)(2a;'  +  3a;-5)  =  0. 

12.  ar»  +  a;'-4(a;  +  l)  =  0. 

13.  3ar»  +  2a;'-(3a;-f 2)  =  0. 

14.  a;«-27-13a;+39  =  0. 

15.  a;^+8  +  3(a;'-4)  =  0.      17.  2ar'-2a;'-(a;»-l) 

16.  a;(a;'-l)-6(a;-l)  =  0.    18.  a;=»  -  3a-2  =  0. 

19.  2ar»  +  2a;'  +  (^*-5a;-6)  =  0. 

20.  a;*-4ar»-ar^  + 16a;- 12  =  0. 


266  SCHOOL   ALGEBRA. 


Solutions  by  a  Formula. 

287.  Every  quadratic  equation  can  be  made  to  assume 
the  form  ax^  -\-hx-\-c  =  0. 

Solving  this  equation  (§  285,  Ex.  1),  we  obtain  for  its 
two  roots 


-&  +  VF^=T^        -5-V6^-4 


ac 


2a  2a 

There  are  two  roots,  and  but  two  roots,  since  there  are 
two,  and  but  two,  square  roots  of  the  expression  b^  —  ^ac. 

By  this  formula,  the  values  of  x  in  an  equation  of  the 
form  aoi^  -\-hx-\-c  =  ()  may  be  written  at  once. 

Ex.   Find  the  roots  of  the  equation  3ar^— 5a;  +  2  =  0. 

Here  a  =  3,  6  =  —  5,  c  =  2. 

Putting  these  values  for  the  letters  in  the  above  formulas,  we  have 

„     5  +  V25  -  24   ^^  5  -  \/25  -  24 
X .  or  -, , 

6  o       <  } 

-forf 
-1  or  |. 

Exercise  97. 

Solve  by  the  above  formulas : 

1.  2a;»  +  3a;=14.  7.  bo(^-*Jx  =  -2. 

2.  3a;' -5a:  =12.  8.  4a;' -9a;  =  28. 

3.  a;'-7a;=18.  9.  5a;'  +  7a;  =  12. 

4.  5a;«-a;  =  42.  10.  lla;»- 9a;  =  -^- 

5.  6a;»-7a;=:10.  11.  7a;' +  5a;  =  38. 

6.  3a;»-lliP  =  -6.  12.  5ar'-7a7  =  6. 


or  THT  ,,  --^ 

UNIVERSITY   J  QUADRATIC  equations.  267 

^      or 
£iL'roRl_ 

tUATIONS   IN    THE    QUADRATIC   FORM. 

288.  An  equation  is  in  the  quadratic  for  mn  if  it  contains 
but  two  powers  of  the  unknown  number,  and  the  exponent 
of  one  power  is  exactly  twice  that  of  the  other  power. 

289.  Equations  not  of  the  second  degree,  but  of  the 
quadratic  form,  may  be  solved  by  completing  the  square. 

(1)  Solve:  8a:«+63a:'  =  8. 

This  equation  is  in  the  quadratic  form  if  we  regard  a^  as  the  un- 
known number. 

We  have  8a;«  +  63a^  =  8. 

Multiply  by  32  and  complete  the  square, 

256  a;«  +  ()  + (63)2  =  4225. 

Extract  the  square  root,     16  cc^  +  63  =  ±  65. 

Hence,  a:^  =  -  or  —  8. 

8 

Extracting  the  cube  root,  two  values  of  x  are  \  and  —  2.  There 
are  four  other  values  of  a;  which  we  do  not  find  at  present. 

(2)  Solve:  V?-3</P^-40. 

Using  fractional  exponents,  we  have 

a;3_  3x^  =  40. 

This  equation  is  in  the  quadratic  form  if  we  regard  x^  as  the  un- 
known number. 

Complete  the  square,  4  a;'  — 12  x^  +  9  =  169. 
Extract  the  root,  2  x^  -  3  =  ±  13. 

.-.  2  a;*  =  16  or -10, 
a;i  =  8or-5, 
a;  =  16  or  5\/5. 
There  are  other  values  of  ar  which  we  do  not  find  at  present 


268 


SCHOOL    ALGEBRA. 


(3)    Solve  completely  the  equation  a;'  =  1. 

We  have  a^  —  1  =•  0. 

Factoring,      {x  —  l){a^  +  x  +  l)='0. 


or 


Therefore,  either 
a?-l  =  0. 


The  three  values,  1, 
roots  of  1. 


a;- 1  =  0 
a*  +  a?  +  1  =  0. 


Solving,    X 


1  + V^   _1_  VITS 


a;*  +  a;  +1  =  0. 
-IdbV^Ts 


,  are. the  three  cube 


(4)   Solve:         (2x-Sy--{2x-S)^6. 
Put  y  for  2  a?  —  3,  and  therefore  y^  for  (2a;  —  3)''. 
We  have  y^  —y  =  Q. 

Solving,  y  =  3  or  —  2. 

Putting  now  2  a;  —  3  for  y, 

2a;- 3  =  3,  I  2a; -3 2, 

a;  =•  3.  I  SB"^. 

Exercise  98. 

1.  a7*-5a;2  +  4  =  0.  6.    10a7*-21  =a:». 

2.  a;*-13a:^+36-0.  -7.    -v/?+ 3-v^=  If. 

3.  x^  -  21  a;'  =  100.  X  8.    Z</x  -  2 V^  =  -  20. 

4.  4a;«~3a;'  =  27.  9.    5 a;'~  +  3 a;"  =  6f . 

5.  2a:*  +  5a;'  =  21|.  10.    (8a:-f3)^+(8a:+3)  =  30. 


11.   2(a;'  — ^7+1)  — Va;'-a7+l  =  l. 
12.   a;«-9ar'  +  8  =  0.  14.    (a;+ 1)  + Va;+ 1  =  6. 

15.   a;*- 13a;' =  —  36. 


13.  .^+.-^=15, 

6 


16.    2a:'  +  4a:  +  9  +  3V2a72  +  4a;  +  9  =  40. 

-«     12a;»-lla;^+10a:-78_.,,     _1 
8a;' -7a; +6  ~  ^'^'^     2 


QUADRATIC   EQUATIONS.  269 


Radical  Equations. 

290.   If  an  equation  involves  radical  expressions,  we  first 

clear  of  radicals  as  follows  : 

Solve  Va7  +  4  +  V2a;  +  6  =  V7a;4-14. 

Square  both  sides, 

a?  +  4  +  2y/{x  +  4)  {2^T^  +  2ar  +  6  =  7a;  +  14. 

Transpose  and  combine,        2V(aj  +  4)(2a;  +  6)  =.  4  a?  +  4. 

Divide  by  2  and  square,  (a;  +  4) (2a;  +  6)  =-  (2a;  +  2)*. 

Reduce,  a;"  — 3  a;  =10. 

Hence,  a;  =  5  or  —  2. 

Of  these  two  values,  only  5  will  satisfy  the  original  equation. 

The  value  —  2  will  satisfy  the  equation 

Va;  +  4  -  V2a;  +  6  =  V7a;  +  14. 

In  fact,  squaring  both  members  of  the  original  equation  is  equiva- 
lent to  transposing  V7a;  +  14  to  the  left  member,  and  then  multiplying 
by  the  rationalizing  factor  Va;+4  +  V2a;  +  6  +  V7a;  +  14,  so  that 
the  equation  stands 

(Va;  +  4  +  V2a;  +  6 -  V7a;  + 14)(V'a;  +  4  +  \/2x-{-Q  +  V7a;+14)  =  0, 
which  reduces  to  V(a;  +  4)(2a;  +  6)  —  (2  a;  +  2)  =■  0. 

Transposing  and  squaring  again  is  equivalent  to  multiplying  by 
(V^T4  -  V2a;  +  6  +  V7a;  +  14)(v^T4-  V2a;  +  6  -  V7a;+14). 

Multiplying  out  and  reducing,  we  have 
a;' -3a; -10-0. 

Therefore,  the  equation  a;'  —  3  a;  —  10  =•  0  is  really  obtained  from 
(Va;  +  4  +  V2a;  +  6-  V7a;  +  14) 


X  (va;  +  4  +  \/2a;  +  6  +  V7a;  +  14) 
X  (Va;  +  4 -  V2a;  +  6-  V7a;  +  14) 
X(Va;+4- V2a;  +  6  +  V7a;  +  14)  =  0. 

This  equation  is  satisfied  by  any  value  that  will  satisfy  any  one 
of  the  four  factors  of  its  left  member.  The  first  factor  is  satisfied 
by  5,  and  the  last  factor  by  —  2,  while  no  values  can  be  found  to 
satisfy  the  second  or  third  factor. 


270  SCHOOL    ALGEBRA. 

Hence,  if  a  radical  equation  of  this  form  is  proposed  for  solution, 
if  there  is  a  value  of  a;  that- will  satisfy  the  particular  equation  given, 
that  value  must  be  retained,  and  any  value  that  does  not  satisfy  the 
equation  given  must  be  rejected.  (See  Wentworth,  McLellan  and 
Glashan's  Algebraic  Analysis,  pp.  278-281.) 

29L   Some  radical  equations  may  be  solved  as  follows : 


Solve  Ix"  -  5a;  +  8V7:r"'  -  5:r  +  1  =  -  8. 
Add  1  to  both  sides, 

7a;2-5a;  +  l  +  8\/7a;2- 5a;  +  1  =  -  7. 
Put  V7  a;"'*  —  5  a;  +  1  =  y  ;  the  equation  becomes 

Hence,  y  =  —  1  or  —  7, 

3/2  =  1  or  49. 
"We  now  have  7a;2  —  5a;  +  l  =  l,  or  7a;2  —  5a;+l==49. 
Solving  these,  we  find  for  the  values  of  x, 

7   17 

These  values  all  satisfy  the  given  equation  when  we  take  the 
negative  value  of  the  square  root  of  the  expression  7x'^  — 5a;  +  l; 
they  are  in  fact  the  four  roots  of  the  biquadratic  obtained  by  clear- 
ing the  given  equation  of  radicals. 


Solve : 


Exercise  99. 

1.  V9a7  +  40  —  2V^TT=  Vx. 

2.  -y/a-i-  X-]-  ■\/a  —  x  =  -Vb. 


3  a:  +  -\/4  x  —  x^      o 

3.    -=  =  ^. 

Sx—  ■v4a;  — a;' 


4.  Vx  —  S--Vx-14:  =  V4a7  —  155. 

5.  ■\/x-j-4i  —  -\/x  =  -\/x-\-^. 


QUADRATIC   EQUATIONS.  271 

'    2  +  Vx       40 +V^ 


7.  Vl4a;  +  9  +  2 Vo; 4-l  +  V3a;+l  =  0. 

8.  V5a7  +  1  -  2  -  VF+l  =  0. 


9.    V^  —  2  +  Vo;  +  3  —  V4 a;  +  1  =  0. 


10.  V7-a;+ V3a:+10+ Va;  +  3  =  0. 

11.  3Va:='  +  17  +  V^Tl  +  2 V5a;»  +  41  =  0. 


12.  2x-'V2x  —  l=x  +  2. 

13.  V:r  +  2  -  V:r  -  2  —  V2^=  0. 

1  ,  1 


14. 


f -12. 


a;  +  V^'^  —  1      ^  —  V:^"'^  —  1 
91 


15.    V3^+ V3a:+13 


V3a;+13 


16.    :r- W-2^'^-2  =  0. 


17.    ,    =x+  -Vx"  —  8. 

X  -  Vx"  -  8 


18.    2a;'  +  3a:-5V2x'^  +  3a;  +  9  =  -3. 


19.  3:r'+15a;-2Va;'-[-5a;+l  =  2. 

20.  x'-^x-i-  SV2a^  -  3a;T2  =  7. 


21.  2x'  —  Va;'  -  2a;  —  3  =  4a;  +  9. 

22.  Sx'  —  4a:  +  V3a;'  -  4a;  -  6  =  18. 

23.  3a;' -  7  +  3V3a;'  —  16a;  +  21  =  16a;. 


272  SCHOOL    ALGEBRA. 

292.  Problems  involving  Quadratics.  Problems  which  in- 
volve quadratic  equations  apparently  have  two  solutions, 
since  a  quadratic  equation  has  two  roots.  When  both  roots 
of  the  quadratic  equation  are  positive  integers,  they  will, 
generally,  both  be  admissible  solutions. 

Fractional  and  negative  roots  will  in  some  problems  give 
admissible  solutions ;  in  other  problems  they  will  not  give 
admissible  solutions. 

No  difficulty  will  be  found  in  selecting  the  result  which 
belongs  to  the  particular  problem  we  are  solving.  Some- 
times, by  a  change  in  the  statement  of  the  problem,  we  may 
form  a  new  problem  which  corresponds  to  the  result  that 
was  inapplicable  to  the  original  problem. 

Imaginary  roots  indicate  that  the  problem  is  impossible. 

Here  as  in  simple  equations  x  stands  for  an  unknown 
number. 

(1)  The  sum  of  the  squares  of  two  consecutive  numbers 
is  481.     Find  the  numbers. 


Let 

X  ■=■  one  number, 

and 

a;  +  1  =  the  other. 

Then 

a;2  +  (a;  +  1)2  =  481, 

or 

2rc2  + 2a; +  1=481. 

The  solution  of  which  gives  a;  =-  15  or  — 16. 

The  positive  root  15  gives  for  the  numbers,  15  and  16. 

The  negative  root  — 16  is  inapplicable  to  the  problem,  as  consecu- 
Inve  numbers  are  understood  to  be  integers  which  follow  one  another 
in  the  common  scale,  1,  2,  3,  4 

(2)  A  pedler  bought  a  number  of  knives  for  $2.40. 
Had  he  bought  4  more  for  the  same  money,  he  would  have 
paid  3  cents  less  for  each.  How  many  knives  did  he  buy, 
and  what  did  he  pay  for  each  ? 

Let  X  =  number  of  knives  he  bought. 

Then  =.  number  of  cents  he  paid  for  each. 


QUADRATIC  EQUATIONS.  273 

Bat  if  a;  +  4  =  number  of  knives  he  bought, 

=  number  of  cents  he  paid  for  each, 

a;  +  4 

= =  the  difference  in  price. 

X       a;  +  4 

But  3  =  the  difference  in  price. 

•   240  240  _^3 

X  a;  +4 

Solving,  a;  =  16  or  -  20. 

He  bought  16  knives,  therefore,  and  paid  ^^,  or  15 
cents  for  each. 

If  the  problem  is  changed  so  as  to  read :  A  pedler  bought 
a  number  of  knives  for  $2.40,  and  if  he  had  bought  4  less 
for  the  same  money,  he  would  have  paid  3  cents  more  for 
each,  the  equation  will  be 

240      240  _  3 
a;— 4       X 
Solving,  a;  =  20  or  -  16. 

This  second  problem  is  therefore  the  one  which  the  neg- 
ative answer  of  the  first  problem  suggests. 

(3)   What  is  the  price  of  eggs  per  dozen  when  2  more  in 
a  shilling's  worth  lowers  the  price  1  penny  per  dozen  ? 
Let  X  ==  number  of  eggs  for  a  shilling. 

Then  -  =  cost  of  1  egg  in  shillings, 

12 
and  —  =•  cost  of  1  dozen  in  shillings. 

X 

But  if  a?  +  2  =•  number  of  eggs  for  a  shilling, 

==  cost  of  1  dozen  in  shillings, 
a;  +  2 

12        12         1 

.*. =-  —  (1  penny  being  ^^  of  a  shilling). 

X      a;  +  2     12 

The  solution  of  which  gives  x  =>  16,  or  —  18. 

And,  if  16  eggs  cost  a  shilling,  1  dozen  will  cost  9  pence. 

Therefore,  the  price  of  the  eggs  is  9  pence  per  dozen. 


274  SCHOOL   ALGEBRA. 

If  the  problem  is  changed  so  as  to  read :  What  is  the 
price  of  eggs  per  dozen  when  two  less  in  a  shilling's  worth 
raises  the  price  1  penny  per  dozen  ?  the  equation  will  be 

12       12 _  1 
x-2      X      12' 

The  solution  of  which  gives  x  =  18,  or  — 16. 

Hence,  the  number  18,  which  had  a  negative  sign  and  was  inappli- 
cable in  the  original  problem,  is  here  the  true  result. 

Exercise  100. 

1.  The  sum  of  the  squares  of  two  consecutive  integers  is 
761.     Find  the  numbers. 

2.  The  sum  of  the  squares  of  two  consecutive  numbers  ex- 
ceeds the  product  of  the  numbers  by  13.    Find  the  numbers. 

3.  The  square  of  the  sum  of  two  consecutive  even  num- 
bers exceeds  the  sum  of  their  squares  by  336.  Find  the 
numbers. 

4.  Twice  the  product  of  two  consecutive  numbers  ex- 
ceeds the  sum  of  the  numbers  by  49.     Find  the  numbers. 

5.  The  sum  of  the  squares  of  three  consecutive  numbers 
is  110.     Find  the  numbers. 

6.  The  difference  of  the  cubes  of  two  successive  odd 
numbers  is  602.     Find  the  numbers. 

7.  The  length  of  a  rectangular  field  exceeds  its  breadth 
by  2  rods.  If  the  length  and  breadth  of  the  field  were 
each  increased  by  4  rods,  the  area  would  be  80  square  rods. 
Find  the  dimensions  of  the  field. 

8.  The  area  of  a  square  may  be  doubled  by  increasing 
its  length  by  10  feet  and  its  breadth  by  8  feet.  Determine 
its  side. 


QUADRATIC   EQUATIONS.  275 

9.  A  grass  plot  12  yards  long  and  9  yards  wide  has  a 
path  around  it.  The  area  of  the  path  is  -J  of  the  area  of  the 
plot.     Find  the  width  of  the  path. 

10.  The  perimeter  of  a  rectangular  field  is  60  rods.  Its 
area  is  200  square  rods.     Find  its  dimensions. 

11.  The  length  of  a  rectangular  plot  is  10  rods  more 
than  twice  its  width,  and  the  length  of  a  diagonal  of  the 
plot  is  25  rods.     What  are  the  dimensions  of  the  field  ? 

12.  The  denominator  of  a  certain  fraction  exceeds  the 
numerator  by  3.  If  both  numerator  and  denominator  be 
increased  by  4,  the  fraction  will  be  increased  by  |.  Deter- 
mine the  fraction. 

13.  The  numerator  of  a  fraction  exceeds  twice  the  de- 
nominator by  1.  If  the  numerator  be  decreased  by  3,  and 
the  denominator  increased  by  3,  the  resulting  fraction  will 
be  the  reciprocal  of  the  given  fraction.     Find  the  fraction. 

14.  A  farmer  sold  a  number  of  sheep  for  $120.  If  he 
had  sold  5  less  for  the  same  money,  he  would  have  received 
$2  more  per  sheep.     How  much  did  he  receive  per  sheep? 

State  the  problem  to  which  the  negative  solution  applies. 

15.  A  merchant  sold  a  certain  number  of  yards  of  silk 
for  $40.50.  If  he  had  sold  9  yards  more  for  the  same 
money,  he  would  have  received  75  cents  less  per  yard. 
How  many  yards  did  he  sell? 

16.  A  man  bought  a  number  of  geese  for  $27.  He  sold 
all  but  2  for  $25,  thus  gaining  25  cents  on  each  goose  sold. 
How  many  geese  did  he  buy  ? 

17.  A  man  agrees  to  do  a  piece  of  work  for  $48.  It 
takes  him  4  days  longer  than  he  expected,  and  he  finds 
that  he  has  earned  $1  less  per  day  than  he  expected  In 
how  many  days  did  he  expect  to  do  the  work  ? 


276  SCHOOL   ALGEBRA. 

18.  Find  the  price  of  eggs  per  dozen  when  10  more  in 
one  dollar's  worth  lowers  the  price  4  cents  a  dozen. 

19.  A  man  sold  a  horse  for  $  171,  and  gained  as  many 
per  cent  on  the  sale  as  the  horse  cost  dollars.  How  much 
did  the  horse  cost  ? 

20.  A  drover  bought  a  certain  number  of  sheep  for  $  160. 
He  kept  4,  and  sold  the  remainder  for  $10.60  per  head,  and 
made  on  his  investment  f  as  many  per  cent  as  he  paid  dollars 
for  each  sheep  bought.     How  many  sheep  did  he  buy  ? 

21.  Two  pipes  running  together  can  fill  a  cistern  in  5| 
hours.  The  larger  pipe  will  fill  the  cistern  in  4  hours  less 
time  than  the  smaller.  How  long  will  it  take  each  pipe 
running  alone  to  fill  the  cistern  ? 

22.  A  and  B  can  do  a  piece  of  work  together  in  18  days, 
and  it  takes  B  15  days  longer  to  do  it  alone  than  it  does  A. 
In  how  many  days  can  each  do  it  alone  ? 

23.  A  boat's  crew  row  4  miles  down  a  river  and  back 
again  in  1  hour  and  30  minutes.  Their  rate  in  still  water 
is  2  miles  an  hour  faster  than  twice  the  rate  of  the  current. 
Find  the  rate  of  the  crew  and  the  rate  of  the  current. 

24.  A  number  is  formed  by  two  digits.  The  units'  digit 
is  2  more  than  the  square  of  half  the  tens'  digit,  and  if  18 
be  added  to  the  number,  the  order  of  the  digits  will  be 
reversed.     Find  the  number. 

25.  A  circular  grass  plot  is  surrounded  by  a  path  of  a 
uniform  width  of  3  feet.  The  area  of  the  path  is  J  the  area 
of  the  plot.     Find  the  radius  of  the  plot. 

26.  If  a  carriage  wheel  11  feet  round  took  -^  of  a  second 
less  to  revolve,  the  rate  of  the  carriage  would  be  5  miles 
more  per  hour.     At  what  rate  is  the  carriage  travelling  ? 


^  CHAPTER  XIX. 


ir 


SIMULTANEOUS    QUADRATIC    EQUATI03SFS. 

293.  Quadratic  equations  involving  two  unknown  num- 
bers require  different  methods  for  their  solution,  according 
to  the  form  of  the  equations. 

Case  I. 

294.  When  from  one  of  the  equations  the  value  of  one  of  the 
miknown  numbers  can  be  found  in  terms  of  the  other,  and  this 
value  substituted  in  the  other  equation. 

Ex.   Solve:         Sx^-2x7/  =  6\     '  (1) 

ar-y  =  2  J  (2) 

Transpose  x  in  (2),  y  —  a;  —  2. 

In  (1)  put  a;  ~  2  for  y, 

3a;2-2a;(a;-2)  =  5, 
The  solution  of  which  gives  a;  —  1,  or  a;  ==  --  5. 
If  a;-l, 

y  =  l_2  =  -l; 
and  if  ,      a;  =  —  5, 

y=,_5-2^ 7. 

We  have  therefore  the  pairs  of  values, 

*-M;   or^  — n. 
y»^li         y--7i 

The  original  equations  are  both  satisfied  by  either  pair  of  values. 
But  the  values  a;  =  1,  y  =  —  7,  will  not  satisfy  the  equations ;  nor  will 
the  values  a;  —  —  5,  y  =■  —  1. 

The  student  must  be  careful  to  join  to  each  value  of  x 
the  corresponding  value  of  y. 


278 


SCHOOL   ALGEBRA. 


Case  II. 

295.  When  the  left  side  of  each  of  the  two  equations  is 
homogeneous  and  of  the  second  degree. 


Solve 


2y»-4a;y  +  3a;'=17 


y»-ar»=16  ) 


Let  y  =  vx,  and  substitute  vx  for  y  in  both  equations. 
From  (1),  2vV  -  4  w'  +  3  a;*  =»  17. 

.-.  x^ 


17 


From  (2), 


Equate  the  values  of  a^, 


2v2-4v  +  3 
r;»a^-a;2=16. 
16 


.-.  a^ 


17 


v2_l 
16 


2i;2-4v  +  3     v2_i 

32v2-64i;  +  48  =  17t;2-17, 

15v2-64v  =  -65, 

225^2 -960v  =  - 975, 

225^2 -0  + (32)2  =  49, 

152;-32  =  ±7. 


If 

v  = 

5 
'3 

y- 

"VX-' 

3  * 

Substitute  in 

(2). 

25a;2 
9 

-a;2=. 

■16, 

a;*  = 

=  9, 

«» 

.±3, 

y- 

5a; 
'  3  " 

-±5. 

-1 

-f 

If 

v  = 

^13 

y 

=  t;a;  = 

13a; 

Substitute  in  (2), 

w 

169^2 
25 

-a:2  = 

^2  = 

=  16, 

25 
'  9' 

a;  = 

-1 

13  ar 

.13 

a) 

(2) 


y- 


SIMULTANEOUS   QUADRATIC   EQUATIONS.  279 

Case  III. 

296.  When  tlie  two  equations  are  symmetrical  with  respect  to 
X  and  y ;  that  is,  when  x  and  y  are  similarly  involved. 

Thus,  the  expressions 

2:r»+3a;»y'+2y',  2xy-Zx-Zy^l,  x'-Zx'y--Zxf-iry\ 

are  symmetrical  expressions.  In  this  case  the  general  rule 
is  to  combine  the  equations  in  such  a  manner  as  to  remove 
the  highest  powers  of  x  and  y. 

Solve:  x'  +  y'  =  2>2>l\  (1) 

a;4-y=     73  (2) 

To  remove  a;*  and  y*,  raise  (2)  to  the  fourth  power, 

x*  +  4ar'3/  +  6a;V  +  4a;/+    y*  =  2401 

Add  (1),       J^ +3/*-    337 

2a;*  +  4a;3y  +  63.2^2  j^^xy^  ^-  2/  =  2738 

Divide  by  2,  x*  +  2  a^y  +  3  x^  +  2  x/  +    y*  =  1369. 

Extract  the  square  root,  a;^  +  a;y  +  y^  =  ±  37.  (3) 

Subtract  (3)  from  (2)2,      xy  =  12  or  86. 

We  now  have  to  solve  the  two  pairs  of  equations, 

a;  +  y=    7).    x  +  y^    71 
/  '         an/  =  86  / 


From  the  first,  ^  =  ^\:     or     ^^^ 

y 


xy  =  12  J  xy 

-=n;  or  --n 

v  =  3i  v=-4/ 


From  the  second. 


7± 

V- 

295 

7t 

2 

V- 

295 

y  - 

2 

. 

280  SCHOOL   ALGEBRA. 

297,  The  preceding  cases  are  general  methods  for  the 
solution  of  equations  which  belong  to  the  kinds  referred  to; 
often,  however,  in  the  solution  of  these  and  other  kinds  of 
simultaneous  equations  involving  quadratics,  a  little  inge- 
nuity will  suggest  some  step  by  which  the  roots  may  be 
found  more  easily  than  by  the  general  method. 


(1)   Solve:                  ar  +  y  =  40" 

(1) 

(2) 

Square  (1),              x^  +  2xy  +  y'^=^  1600. 

(3) 

Multiply  (2)  by  4,                 4  xy  =  1200. 

(4) 

Subtract  (4)  from  (3), 

a;2_2a;y+y'  =  400. 

(5) 

Extract  root  of  each  side,  a;  —  y  =  ±  20. 

(6) 

From  (1)  and  (6),                       a;  =  30  ■)  . 

y  =  ior 

(2)    Solve:                  1  +  1=  A 
^  ^                                  X       y      20 

(1) 

1,1       41 
s^'^f     400  J 

(2) 

Square  (1),                 1  +  A  +  1  =  il . 
x^     xy     3/2     400 

(3) 

Subtract  (2)  from  (3),               ~-^ 
xy     400 

(4) 

Subtract  (4)  from  (2), 

1       2       1        1 
x^     xy     3/2     400 

Extract  the  root,                 1  _  2  =  ±  .1. 
X     y        20 

(5) 

From  (1)  and  (5),                       a;  =  4 1     ^ 
3/  =  5/' 

y==4J 

SIMULTANEOUS    QUADRATIC   EQUATIONS.  281 

(3)   Solve:                 a;  -y  =    4"!  *  (1) 

x'-\-y  =  4:0f  (2) 

Square  (1),              x^ -2xi/ +  y^  =  ie.  '                          (3) 

Subtract  (2)  from  (3),         -  2  xy  =  -  24.  (4) 
Subtract  (4)  from  (2), 

a;2  +  2xy  +y^  =  64:. 

Extract  the  root,                 x  +  y  =  ±8.  (6) 
From  (1)  and  (5), 


y=2/'       y  =  -6/ 


(4)   Solve:  .  ar«  +  y'  =  91'l  (1) 

X  +y  =    7)  (2) 

Divide  (1)  by  (2),     x^-xy+y^=  13.  (3) 

Square  (2),  x^  +  2xy+y^  =  49.  (4) 

Subtract  (3)  from  (4),  3xy  =  36. 

Divide  by  -  3,  -xy  =  - 12.  (5) 

Add  (5)  and  (3),     x'-2xy+y^  =  l. 

Extract  the  root,  x  —  y  =  ±l.  (6) 

From  (2)  and  (6),  »  =  4j,  ^^a;  =  3| 


(5)   Solve:  af'  +  f  =18x^1  (1) 

a:+y=12      J  (2) 

Divide  (1)  by  (2),     x^-xy+y^  =  ^.  (3) 

Square  (2),  x^  +  2xy+y^=^  144.  (4) 

Subtract  (4)  from  (3),         _  3  xy  =  ^  -  144, 

which  gives  xy  =  32. 

We  now  have,  ar  +  y  =  12 1 

a:y  =  32  /  " 

Solving,  we  find,  x=81     ^^a;  =  4) 

y=.4/'       y-si" 


282 


SCHOOL    ALGEBRA. 


Exercise  ±01. 


-10     J 


xy 


y=Q 


xy 


] 


5.    X 

x' 


x-j-y  =  l2')        4.    x-y=10') 
xy  =  27       I  xy=ll       ) 

! 


xy 

7.  X  —y  =    9 
x'  +  y'  =  ib 

8.  X  +">■ 


+  3/^  =  80] 
:i:-^  -f  ,y2  ^  29  ) 


7) 

10  I 


9.    3a;  -y  =12 


y 


=  16 


10.    y  =  3a;+l 
x^J^xy  =  33 


11.  5.r  -4y  =10) 
3a;-^--4y'=    8) 

12.  a;-f  7y  =  23) 
a:y  =  6  j 

13.  2a;  -3y    =2     | 

x'-lxy^-l] 

14.  2.r  -3y    =    1) 
3a;^-4a;y  =  32) 

15.  :c'-a;y+y'  =  21| 
a;  +  y  =  9  3 

16.  x^ 

2a;  +  3y=: 

17.  x'  — ?/'  = 


3a:y  +  2y'  =  0| 


X    —  2/    =  1  ) 


18.    a;'-f  3y  +  17=0) 
Zx-y  =  Z  ) 

1 


19. 


20. 


-[--  =  5 

^    y 
1-1  =  5. 

X     y     ^ 

2.  ,1^13 

x^     y"     36 


21.  2  +  5  =  2 

a;y  =  6 

22.  i+i=ll 
X     y 

3  +  ^  =  61 
^    y' 

23.  5+8^4' 

^    y 


a:y  =  16 


24 


^  +y  =  5j 


SIMULTANEOUS    QUADRATIC    EQUATIONS. 


283 


25.  r'-y'  =  61) 
ar  -y  =    1) 

26.  a^  +  y^  =  65) 
X  +7/  =:    5) 

27.  x'^-{-xi/^  =  l20) 

28.  x'-y'  =  ^^) 

^  -y  =  V 

29.  a;'  +  y»  =  126         ") 

^'-^y  +  y'  =  2if 

30.  x^-y'^be  I 
2/^  =  28  [ 


31. 


32. 


x'  +  xi/-\- 

x^^     12 

y      a;  ~  2 
y     a?~18 


33.  a:'  +a;y  =24" 
^y  +   y'  =  40 . 

34.  a:"  -  a:y  =  8 1 
a;y_   y«=7l 


35.   a^H-2ry=:24 


2a:y +  4y* 

4  a:'  +  5  a:i 
7a:3/  +  9y 


::24         ) 

=  120f 


36.    4a:'  +5a:y  =  14) 
=  50  1 


37.  a:'  +  a:y  +  y'  =  39       ) 
2a:'  +  3a:y  +  y2  =  63) 

38.  a;'  +3y'  =  52) 
a:y  +  2y'^  =  40) 

39.  2a;'-y'  =  46| 

40.  x'  +  xy  +  27/^==U     ) 
2x'-Sxyi-2y'=l6\ 

41.  a;2  +  33/'  =  31  | 
4a:y  +  3/^  =  33) 

42.  3a:'+7a7y  =  82  ) 
x'-{-5xy  +  9f  =  219i 

43.  a;*  +  y*  =  97) 
X  -\-y  =   5  ) 

44.  a:*  +  3/*  =  17) 
a:  +y  =    3j 

45.  a;*+y*  =  881) 
X  —y  —      1  ) 

46.  a:*-l-y*  =  211) 
a:  +y  =      if 

47.  a:*  — y*  =  242 


y 


=  242) 
=      2) 


48.  a;^  +  y>  =  a:y+7) 
X  -\-y  -=xy~l) 

49.  ar*  — y*=7a:y") 
ar  -y  =2      ) 


2S4 


SCHOOL 


BO.    a^  +  f  =  Se>xy)^ 
a;  +y  =24       J 


ALGEBRA 

61.   ar' 


61.   x'  +  Sx7/'  =  62 

3; 


52.    a:'  +  a:y  +  y'  =      61 1 


53.    x^-  XT/  +1/^^    3 
^*  +  ^y  +  3/*  =  21 


^*  +  a7y  +  y*  =  l28l 

1 

a?  —  y     a?  +  y_24 
x-\-y~x  —  y~  5 

3a;  +  4y  =  36 


56.   a;''  +  y'  +  a;  +  y  =  32| 
a;y+16  =  0  j 


54. 


65. 


67 


.   a?  — y  — 3  =  0      ) 

2(a;'-2/')  =  3a;yj 


68.    1  +  1  =  7 
a;     y 


1 


1 


31 


27+1     y+1      20^ 

69.    a:*+S^*  =  272         | 
a;*  +  y'  =  3a:y-4j 

60.   a;'  +  y'  =  ar'y'+l| 
X  +y  —2xy—l) 


x  — y  =a  ) 


62.    ?  +  f=l 
a      h 

^  +  ^  =  4 
^     y 


63.  x^  =  ax-\-hy\ 
y^^hx-^ay] 

64.  a;'  +  y'  =  2(a*  +  *')| 


65.    a;'  +  y'  =  ^l±A* 
xy=l 

a  —  h' 


66.   x^  —  y' 


a  +  b 


xy 


ah 


{a+by 


67.  x"  —xy  =2a^>  +  25M 
a;y-   y'' =  2  aZ>  -  2  ^)M 

68.  x'  —  y''  =  a^ 


xy 


3 


69.  a;^~y'  =  8a5| 
a;y  =  a*  —  4  i''  J 

70.  ar»  +  y»  =  a»+5') 
X  +y  =o  +^>  ) 


SIMULTANEOUS   QUADRATIC   EQUATIONS.  285 


Exercise  102. 

1.  The  area  of  a  rectangle  is  60  square  feet,  and  its 
perimeter  is  34  feet.  Find  the  length  and  breadth  of  the 
rectangle. 

2.  The  area  of  a  rectangle  is  108  square  feet.  If  the 
length  and  breadth  of  the  rectangle  are  each  increased  by 
3  feet,  the  area  will  be  180  square  feet.  Find  the  length 
and  breadth  of  the  rectangle. 

3.  If  the  length  and  breadth  of  a  rectangular  plot  are 
each  increased  by  10  feet,  the  area  will  be  increased  by  400 
square  feet.  But  if  the  length  and  breadth  are  each  dimin- 
ished by  5  feet,  the  area  will  be  75  square  feet.  Find  the 
length  and  breadth  of  the  plot.  .. 

4.  The  area  of  a  rectangle  is  168  square  feet,  and  the 
length  of  its  diagonal  is  25  feet.  Find  the  length  and 
breadth  of  the  rectangle. 

5.  The  diagonal  of  a  rectangle  is  25  inches.  »If  the 
rectangle  were  4  inches  shorter  and  8  inches  wider,  the 
diagonal  would  still  be  25  inches.  Find  the  area  of 
the  rectangle. 

6.  A  rectangular  field,  containing  180  square  rods,  is 
surrounded  by  a  road  1  rod  wide.  The  area  of  the  road 
is  58  square  rods.     Find  the  dimensions  of  the  field. 

7.  Two  square  gardens  have  a  total  surface  of  2137 
square  yards.  A  rectangular  piece  of  land  whose  dimen- 
sions are  respectively  equal  to  the  sides  of  the  two  squares, 
will  have  1093  square  yards  less  than  the  two  gardens 
united.     What  are  the  sides  of  the  two  squares  ? 

8.  The  sum  of  two  numbers  is  22,  and  the  difference 
of  their  squares  is  44.     Find  the  numbers. 


286  SCHOOL    ALGEBRA. 

9.    The    difference   of    two   numbers   is   6,    and   their 
product  exceeds  their  sum  by  39.     Find  the  numbers. 

10.  The  sum  of  two  numbers  is  equal  to  the  difference 
of  their  squares,  and  the  product  of  the  numbers  exceeds 
twice  their  sum  by  2.     Find  the  numbers. 

11.  The  sum  of  two  numbers  is  20,  and  the  sum  of  their 
cubes  is  2060.     Find  the  numbers. 

12.  The  difference  of  two  numbers  is  5,  and  the  differ- 
ence of  their  cubes  exceeds  the  difference  of  their  squares 
by  1290.     Find  the  numbers. 

13.  A  number  is  formed  of  two  digits.  The  sum  of  the 
squares  of  the  digits  is  58.  If  twelve  times  the  units' 
digit  be  subtracted  from  the  number,  the  order  of  the 
digits  will  be  reversed.     Find  the  number. 

14.  A  number  is  formed  of  three  digits,  the  third  digit 
being  twice  the  sum  of  the  other  two.  The  first  digit  plus 
the  product  of  the  other  two  digits  is  25.  If  180  be  added 
to  the  number,  the  order  of  the  first  and  second  digits  will 
be  reversed.     Find  the  number. 

15.  There  are  two  numbers  formed  of  the  same  two 
digits  in  reverse  orders.  The  sum  of  the  numbers  is  33 
times  the  difference  of  the  two  digits,  and  the  difference  of 
the  squares  of  the  numbers  is  4752.     Find  the  numbers. 

16.  The  sum  of  the  numerator  and  denominator  of  a  cer- 
tain fraction  is  5 ;  and  if  the  numerator  and  denominator 
be  each  increased  by  3,  the  value  of  the  fraction  will  be 
increased  by  ^.     Find  the  fraction. 

17.  The  fore  wheel  of  a  carriage  turns  in  a  mile  132 
times  more  than  the  hind  wheel ;  but  if  the  circumferences 
were  each  increased  by  2  feet,  it  would  turn  only  88  times 
more.     Find  the  circumference  of  each. 


CHAPTER  XX. 
PROPERTIES    OF    QUADRATICS. 

298.  Every  affected  quadratic  can  be  reduced  to  the  form 
ax^  -\-bx-\-c  =  0,  of  which  the  two  roots  are 

h    ^  -^h'-^ac   ^^^_.h ■^Jh''~^ac    §285. 

2a  2a  2a  2a 

Character  of  the  Roots. 

299.  As  regards  the  character  of  the  two  roots,  there  are 
three  cases  to  be  distinguished. 

I.  If  b'^  —  4ac  is  positive  and  not  zero.  In  this  case  the 
roots  are  real  and  unequal.  The  roots  are  real,  since  the 
square  root  of  a  positive  number  can  be  found  exactly  or 
approximately.  If  h"^  —  4  ao  is  a  perfect  square,  the  roots 
are  rational;  if  lP-~^ac  is  not  a  perfect  square,  the  roots 
are  surds. 

The  roots  are  unequal,  since  V<^"''  —  4a<?  is  not  zero. 

II.  If  V  — 4ac  is  zero.     In  this  case  t 
real  and  equals  since  they  both  become  — 


II.   If  V  — 4ac  is  zero.     In  this  case  the  two  roots  are 

2  a 


III.  If  b^  — 4ac  is  negative.  In  this  case  the  roots  are 
iTnaginary,  since  they  both  involve  the  square  root  of  a 
negative  number. 

The  two  imaginary  roots  of  a  quadratic  cannot  be  equal, 
since  H^  —  ^ac  is  not  zero.     They  have,  however,  the  same 


288  SCHOOL   ALGEBRA. 

real  part,  —  -— ,  and  the  same  imaginary  parts,  but  with 

opposite  signs ;  such  expressions  are  called  conjugate  im- 
aginaries.  The  expression  5'  —  4  ac  is  called  the  discriminant 
of  the  expression  aoc^  -\-bx-{-c. 

300.  The  above  cases  may  also  be  distinguished  as  follows : 

Case     I.    i'^  —  4  ac  >  0,  roots  real  and  unequal. 
Case   II.    5^  —  4ac  =  0,  roots  real  and  equal. 
Case  III.    J'  —  4ac  <  0,  roots  imaginary. 

301.  By  calculating  the  value  of  5^  — 4  a*?  we  can  deter- 
mine the  character  of  the  roots  of  a  given  equation  without 
solving  the  equation. 

(1)  a;'  — 5a;  +  6  =  0. 

Here  a  =  1,  6  =  —  5,  c  =  6. 

62_4ac  =  25-24=l. 
The  roots  are  real  and  unequal,  and  rational. 

(2)  ^x^-[-lx-1^0. 

Here  a  =  3,  6  =  7,  c  =  -]. 

&2_4ac  =  49  +  12  =  61. 
The  roots  are  real  and  unequal,  and  are  both  surds. 

(3)  ^x''-l2x  +  ^  =  0. 

Here  a  =  4,   6  =  -  12,  c  =  9. 

J2_4ac  =  144 -144  =  0. 
The  roots  are  real  and  equal. 

(4)  2a;'-3a;  +  4  =  0. 

Here  a  =  2,   6  =  -  3,  c  =  4. 

6»-4ac  =  9-32  =  -23. 
The  roots  are  both  imaginary. 


PROPERTIES   OF   QUADRATICS.  289 

(5)   Find  the  values  of  m  for  which  the  following  equa- 
tion has  its  two  roots  equal : 

2mx'  +  (5m  +  2)a:  +  (4m  +  1)  =  0. 
Here  a  =  2m,  &  =  5m  +  2,  c  =  4m  +  l. 

If  the  roots  are  to  be  equal,  we  must  have 

62  _  4ac  =  0,  or  (5m  +  2)^  -  8m(4m  +  1)  -  0. 

2 
This  gives  m  =  2,  or  —  -. 

For  these  values  of  m  the  equation  becomes 

4a;2  +  12a;  +  9  =  0,  and  4a;2-   4a;  +  l  =  0, 
each  of  which  has  its  roots  equal. 

Exercise  103. 

Determine  without  solving  the  character  of  the  roots  of 
each  of  the  following  equations : 

1.  ^'  +  5a;  +  6  =  0.  6.  6a;' —  7a;  — 3  =  0. 

2.  a;' +  2a;— 15  =  0.  7.  5a;' -  5a;  -  3 -=  0. 

3.  a:'4-2a;  +  3  =  0.  8.  2a;'  — a;  +  5  =  0. 

4.  3a;'  +  7a;  +  2  =  0.  9.  6a;' +  a;- 77  =  0. 

5.  9a;'  +  6a;+l  =  0.  10.    5a;' +  8a;  +  ^  =  0. 

5 

Determine  the  values  of  m  for  which  the  two  roots  of 
each  of  the  following  equations  are  equal : 

11.  (m  +  l)a;'4-(m  — l)a;  +  m  +  l  =  0. 

12.  (2m  — 3)a;'  + ma; +  m  — 1  =  0. 

13.  2ma;'  +  a;'  +  4a;4-2ma;  +  2m  — 4  =  0. 

14.  2mar'  +  3ma;  — 6  =  3a;  — 2m  — a;". 

15.  ma;' +  9a;  -  10  =  3ma;- 2a;' +  2m. 


290  school  algebra. 

Relations  of  Roots  and  Coefficients. 

302.  Consider  the  equation  a:'  —  10  a?  +  24  =  0.  Resolve 
into  factors,  (x  —  6)  (ar  —  4)  =  0.  The  two  values  of  x  are 
6  and  4 ;  their  sum  is  10,  the  coefficient  of  x  with  its  sign 
changed ;  their  product  is  24,  the  third  term. 

303.  In  general,  representing  the  roots  of  the  quadratic 
equation  ax^  -\-bx-{-  g  =  0  by  rj  and  rj,  we  have  (§  285), 

h    ,  -Vb^  —  Aac 
^^^~2^+^ 

r  =-A_ 
Adding,  n  +  r^-- 


h        -s/h'^-^ao 


2a  2a 

a' 


multiplying,  TiT. 


a 


If  we  divide  the  equation  ax^  -\-bx-\-  c  =  0  through  by 

b         c 
a,  we  have   the   equation  x^-{--x-{'-  =  0;    this  may  be 

b  c 

written  x'^ -\- px -{- q  ^=  0  where  p  =  -,  2'  =  — 

It  appears,  then,  that  if  any  quadratic  equation  be  made 
to  assume  the  form  x^-\-px  ~{-  q  =  0,  the  following  relations 
hold  between  the  coefficients  and  roots  of  the  equation  : 

(1)  The  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  X  with  its  sign  changed. 

(2)  The  product  of  t-he  two  roots  is  equal  to  the  constant 
term. 

Thus,  the  sum  of  the  two  roots  of  the  equation 

x'-lx  +  S^O 
is  7,  and  the  product  of  the  roots  8. 


PROPERTIES    OF    QUADRATICS.  291 

304.  Eesolution  into  Pactors.     By  §  303,  if  n  and  r^  are 

the  roots  of  the  equation    x^  -\-px -\-  q  =  0,  the    equation 
may  be  written 

x"  —  (rj  +  r^)x-^r^r^  =  0. 

The  left  member  is  the  product  of  a;  —  ri  and  x  —  rj,  so 
that  the  equation  may  be  also  written 

(x  —  n)  (x  -  r,)  =  0. 

It  appears,  then,  that  the  factors  of  the  quadratic  expres- 
sion x^  ■\-'px-\-  q  are  x—Tx  and  x  —  r^,  where  r^  and  r^  are 
the  roots  of  the  quadratic  equation  x^  -{- px  -\-  q  =^  0. 

The  factors  are  real  and  different,  real  and  alike,  or 
imaginary,  according  as  rj  and  r^  are  real  and  unequal, 
real  and  equal,  or  imaginary. 

If  ra  =  ri,  the  equation  becomes  (x  —  r-^  {x  —  rj  =  0,  or 
(a;  — ri)'  =  0;  if,  then,  the  two  roots  of  a  quadratic  equa- 
tion are  equal,  the  left  member,  when  all  the  terms  are 
transposed  to  that  member,  will  be  a  perfect  square  as 
regards  x. 

305.  If  the  equation  is  in  the  form  aoi^ -\-hx  -\-  c  =  0,  the 
left  member  may  be  written 


a(  x^  -{-  -  X  -\-  -\ 
\        a        aj 

a(x~rx)(x  —  r^). 


306.    If  the  roots  of  a  quadratic  equation  are  given,  we 
can  readily  form  the  equation. 

5 
Ex.    Form  the  equation  of  which  the  roots  are  3  and  —  -• 

The  equation  is       (x  —  3)  f  a;  +  -  J  =  0, 

or  (a;-3)(2a;  +  5)=.0, 

or  2a;2-a?-15  =  0. 


292  SCHOOL    ALGEBRA. 

307.  Any  quadratic  expression  may  be  resolved  into 
factors  by  putting  the  expression  equal  to  zero,  and  solving, 
the  equation  thus  formed. 

(1)    Resolve  into  two  factors  x^  —  bx-{-S. 
Write  the  equation 

Solve  this  equation,  and  the  roots  are  found  to  be 

^^^and^-^. 
2  2 

Therefore,  the  factors  of  a;'^  —  5  a;  +  3  are 

5+  \/l3  „„,         5-\/l3 


X  — 


and  X  — ' 


2  2 

(2)    Resolve  into  factors  3a;^  — 4a;+5. 
Write  the  equation 

3x2-4a;  +  5  =  0. 
Solve  this  equation,  and  the  roots  are  found  to  be 

^^^^^and^-^^^. 
3  3 

Therefore,  the  expression  Sx'^  —  4:x  +  5  may  be  written  (^  305), 

o/      2  +  -\/^Yi\f      2-\/^^rr 


■)H--^} 


Exercise  104. 
Form  the  equations  of  which  the  roots  are 

1.  7,6.  5.    li-li  9.    3  +  V2,  3-V2. 

2.  5,-3.       6.    -11-, -1|.      10.    1  +  V^,  1-V^ 

3.  1^,-2.     7.    13, -4J.  11.    a,a~b. 

4.  4,2^.         8.    A,!..  12.    a  +  b,a-b. 

Resolve  into  factors,  real  or  imaginary : 

13.  12a:'+a;-l.  16.    x'-2x+S. 

14.  3:r'-14a;-24.  17.    ar'  +  x+l. 
16.    :r^-2a:-2.                             18.    ^'  — 2a;  +  9. 


CHAPTER  XXI. 

RATIO,    PROPORTION,    AND    VARIATION. 

308.  Eatio  of  Numbers.  The  relative  magnitude  of  two 
numbers  is  called  their  ratio  when  expressed  by  the  indi- 
cated quotient  of  the  first  hy  the  second.     Thus  the  ratio 

of  a  to  5  is  -,  or  a  -J-  5,  or  a :  b. 

0 

The  first  term  of  a  ratio  is  called  the  antecedent,  and  the 
second  term  the  consequent.  When  the  antecedent  is  equal 
to  the  consequent,  the  ratio  is  called  a  ratio  of  equality; 
when  the  antecedent  is  greater  than  the  consequent,  the 
ratio  is  called  a  ratio  of  greater  inequality ;  when  less,  a  ratio 
of  less  inequality. 

When  the  antecedent  and  consequent  are  interchanged, 
the  resulting  ratio  is  called  the  inverse  of  the  given  ratio. 
Thus,  the  ratio  3  :  6  is  the  inverse  of  the  ratio  6  :  3. 

309.  A  ratio  will  not  be  altered  if  both  its  terms  are 
multiplied  by  the  same  number.      For  the  ratio  a\b  ia 

represented  by  j,  the  ratio  ma :  mb  is  represented  by  — -  ; 
o  mb 

and  since  — -  =  r-,  we  have  ma : mb  =  a:b. 
m,b      b 

310.  A  ratio  will  be  altered  if  different  multipliers  of  its 
terms  are  taken ;  and  will  be  increased  or  diminished  accord- 
ing as  the  multiplier  of  the  antecedent  is  greater  than  or 
less  than  that  of  the  consequent.     Thus, 


294 

SCHOOL   ALGEBRA 

If 

m>n, 

If 

771  <w, 

then 

ma  >  na, 

then 

ma  <  na, 

and 

ma     na . 
nb      nb 

and 

ma     na . 
nb      nb 

but 

na  _a 
nh      h 

but 

na  _a 
nb      h 

.-. 

m>a      a 
nb  ^h' 

•'• 

ma      a 
nb       b 

or 

ma  -.nb^  a:b. 

or 

ma :nh  <a:h. 

311.  Ratios  are  compounded  by  taking  the  product  of  the 
fractions  that  represent  them.  Thus,  the  ratio  compounded 
oi  a:b  and  c:d  is  ac:  bd. 

The  ratio  compounded  of  a :  5  and  a:b  is  the  duplicate 
ratio  a^ :  b^ ;  the  ratio  compounded  of  a:b,  a:b,  and  a:b  is 
the  triplicate  ratio  a^:¥°,  and  so  on. 

312.  Ratios  are  compared  by  comparing  the  fractions 
that  represent  them. 

Thus,  a\h'>  or  <,c  :  d, 

according  as 


as 


as 


a^         ^  G 

ad^  .be 

a(f  >  or  <  be. 


313.  Proportion  of  Numbers.  Four  numbers,  a,b,c,  d,  are 
said  to  be  in  proportion  when  the  ratio  a:h  is  equal  to  the 
ratio  c :  d. 

We  then  write  a:b  =  c:d,  and  read  this,  the  ratio  of  a  to 
b  equals  the  ratio  of  c  to  d,  or  a  is  to  i  as  c  is  to  d. 

A  proportion  is  also  written  a:b::c:d. 

The  four  numbers,  a,  b,  c,  d,  are  called  proportionals;  a 
and  d  are  called  the  extremes,  b  and  c  the  means. 


EATIO    AND    PROPORTION.  295 

314.    When  four  numbers  are  in  proportion,  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 


For,  if  a:h  =  c:df 

n  a      c 

then  -  =  - 

Multiplying  by  bd,        ad=bc. 

The  equation  ad=hc  gives  a  =  ~-,  b  =  — ;  so  that  an 

d  e 

extreme  may  be  found  by  dividing  the  product  of  the 
means  by  the  other  extreme  ;  and  a  mean  may  be  found  by 
dividing  the  product  of  the  extremes  by  the  other  mean. 
If  three  terms  of  a  proportion  are  given,  it  appears  from 
the  above  that  the  fourth  term  has  one  value,  and  but  one 
value. 

315.  If  the  product  of  two  numbers  is  equal  to  the  prod- 
uct of  two  others,  either  two  may  be  made  the  extremes  of 
a  proportion  and  the  other  two  the  means. 

For,  if  ad  —  be, 

then,  dividing  by  5c?,         i-,~-A,' 
od     bd 

or  ^  =  ^. 

b     d 

.*.  a\b  =  c'.d. 

316.  Transformations  of  a  Proportion.  If  four  numbers,  a, 
h,  c,  d,  are  in  proportion,  they  will  be  in  proportion  by : 

I.   Inversion ;  that  is,  b  will  be  to  a  as  c?  is  to  c. 

For,  if  a:b  =  c:d, 

then  —  =  — » 

h     d 


296  SCHOOL   ALGEBRA. 


and  l^2=i^£ 

o  a 

OT  _  =  _. 

a      G 


,\  b  :a  =  d:c. 
11.   Composition ;  that  is,  a-\-h  will  be  to  5  as  c+c?  is  to  d. 
For,  if  ^  a:h  =  Old, 


then 


b~d' 


and  |+l=f+l, 


or 


6  d 

III.   Division ;  that  is,  a  —  5  will  be  to  5  as  c  —  c?  is  to  d. 
For,  if  a:b  =c:df 


then 
and 


or 


a 

b^ 

c 

=5' 

a 
b 

-1  = 

C 

~d 

1, 

a- 

-b 

c  — 

d 

b 

d 

a 

-b'. 

:b  = 

=  c  — 

d: 

d. 

IV.   Composition  and  Division;   that  is,  a-{-b  will  be  to 
a  —  b  as  c  +  c?  is  to  c  —  d. 

For,  from  II.,  24:*  =  ^, 

b  d 


and  from  III., 
Dividing, 


a  —  b     c  —  d 


b  d 

a-\-b  _c-\-d 
a  —  b     c  —  d 
.*.  a-\-b'.a  —  b=c-{-d\c  —  d. 


RATIO   AND    PROPORTION.  297 

V.   Alternation ;  that  is,  a  will  be  to  c  as  5  is  to  d. 
For,  if  a  :  b  =  c  :  d, 

,-,  a     0 

then  T~--; 

b      d 

Multiplying  by  — ,         -—  =  — -. 
^ -^     ^        c  be      cd 

a     b 
or  -  =  -. 

c      a 

.'.  a  :  c  =  b  :  d. 

317.  In  a  series  of  equal  ratios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

For.  if         f  =  S  =  7  =  f' 
b      d    J      h 

r  may  be  put  for  each  of  these  ratios. 

Then  f  ^'^'  ^'^^'  7'^'^' f  "^^^ 

.*.  a  =  br,  c=  dr,  e  =fr,  g  =  hr. 
.-.  a-\-c  +  e-\-g  =  {b  +  d+f-\-h)r, 

" b+d+f+h  b 

.'.  a-{-c-}-e-}-g:b-\-d  +/+  h  =  a:b. 
In  like  manner  it  may  be  shown  that 

ma-\-nc  -j-pe-]-  qg  :  'mb-{-nd-\-j)f-{-qh=a  :  b. 

318.  Continued  Proportion.  Numbers  are  said  to  be  in 
continued  proportion  when  the  first  is  to  the  second  as  the 
second  is  to  the  third,  and  so  on.     Thus,  a,  b,  c,  d,  are  in 

continued  proportion  when  t  —  -  —  ^' 
^  bed 


298  SCHOOL    ALGEBRA. 

319.  If  a,  h,  c  are  proportionals,  so  that  a\h  =  h\c,  then 
h  is  called  a  mean  proportional  between  a  and  c,  and  c  is 
called  a  third  proportional  to  a  and  h. 

If  a:h  =  h'.c,  then  5  =  Vac. 

For,  if  a:h  =  h  :  c, 

then  -  =  -» 

and  h^^ac. 

.'.  h  =  Vac. 

320.  The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 

For,  if  a  :  b  =  c  :  d, 

k  :  I  —  m  :  n, 

then  2  =  -?,  ?  =  2,   ^  =  '^. 

b      d    f     h     I      n 

Taking  the  product  of  the  left  members,  and  also  of  the 
right  members  of  these  equations, 

aek cgm 

bfl      dhn 
.*.  aek  :  bfl  =  cgm  :  dhn. 

321.  Like  powers,  or  like  roots,  of  the  terms  of  a  propor- 
tion are  in  proportion. 

For,  if  a  :  b  =  c  :  d, 

then  ?  =  4. 

b      d 

Raising  both  sides  to  the  nth  power, 

5"  ~  df** 
.-.  a" :  £*»  =  c"  :  d\ 


RATIO   AND   PROPORTION.  299 


Ob  C 

Extracting  the  nth  root,  —  =  —  • 
If      d^ 
L      L        LI 


322.  The  laws  that  have  been  established  for  ratios 
should  be  remembered  when  ratios  are  expressed  in  frac- 
tional form. 

(1)    Solve:         ^;  +  ^+^  =  ^;-^  +  ^. 

By  composition  and  division, 

2x^  2aj2 


2(a;  +  l)      -2(a;-2) 

This  equation  is  satisfied  when  a;  =  0.     For  any  other  value  of  a;, 
we  may  divide  by  x^. 
We  then  have 


x+1      2—x 
and  therefore  a;  =  ^. 

{2)   If  a:b  =  c:d,  show  that 

a^  -{- ab  :  b^  —  ab  =  c^  -}-  cd :  d^  —  cd. 


If 

a       c 
b^d' 

then 

a  +  b     c-^d 
a—b     c—d 

and 

a         c 
-b     -d 

.     a   ^.a  +  b  _    c       c  +  d. 
"  —b'  a  —  b     —d  '  c  —  d' 

that  is. 

a'  +  a6     c'  +  cd 
h^-ab     d^-cd' 

or 

a*  +  a6 :  6'  —  a6  =  <ja  +  cd :  tf*  —  cd 

300  SCHOOL    ALGEBRA. 

(3)    If  a:h  =  c:d,  and  a  is  tlie  greatest  term,  sliow  that 
a  -f  c?  is  greater  than  h-\-c. 


Since 

^  =  3,  and  a>c. 
0     a 

(1) 

the  denominator 

h>d. 

From  (1),  by  division, 

a—h      c—d 
h            d 

(2) 

Since 

h>d, 

from  (2), 

a~b^  c  —  d. 

Now, 
Adding, 

h  +  d=b  +  d, 
a  +  d>b  +  c. 

323.  Eatio  of  Quantities.  To  measure  a  quantity  of  any 
kind  is  to  find  out  how  many  times  it  contains  another 
known  quantity  of  the  same  kind,  called  the  unit  of  measure. 
Thus,  if  a  line  contains  5  times  the  linear  unit  of  measure, 
one  yard,  the  length  of  the  line  is  5  yards. 

324.  Commensurable  Quantities.  If  two  quantities  of  the 
same  kind  are  so  related  that  a  unit  of  measure  can  be 
found  which  is  contained  in  each  of  the  quantities  an  in- 
tegral number  of  times,  this  unit  of  measure  is  a  common 
measure  of  the  two  quantities,  and  the  two  quantities  are 
said  to  be  commensurable. 

If  two  commensurable  quantities  are  measured  by  the 
same  unit,  their  ratio  is  simply  the  ratio  of  the  two  numbers 
by  which  the  quantities  are  expressed.  Thus,  |-  of  a  foot  is 
a  common  measure  of  2J  feet  and  Sf  feet,-  being  contained 
in  the  first  15  times  and  in  the  second  22  times. 

The  ratio  of  2^  feet  to  Sf  feet  is  therefore  the  ratio  of 
15 :  22. 

Evidently  two  quantities  different  in  kind  can  have  no 
ratio. 


i' 


RATIO   AND    PROPORTION.  301 

325,  Incommensurable  Quantities.  The  ratio  of  two  quan- 
tities of  the  same  kind  cannot  always  be  expressed  by  the 
ratio  of  two  whole  numbers.  Thus,  the  side  and  diagonal 
of  a  square  have  no  common  measure ;  for,  if  the  side  is  a 
inches  long,  the  diagonal  will  be  aV2  inches  long,  and  no 
measure  can  be  found  which  will  be  contained  in  each  an 
integral  number  of  times. 

Again,  the  diameter  and  circumference  of  a  circle 
have  no  common  measure,  and  are  therefore  incommen- 
surable. 

In  this  case,  as  there  is  no  common  measure  of  the  two 
quantities,  we  cannot  find  their  ratio  by  the  method  of 
§  324.     We  therefore  proceed  as  follows : 

Suppose  a  and  5  to  be  two  incommensurable  quantities 
of  the  same  hind.  Divide  h  into  any  integral  number  {n) 
of  equal  parts,  and  suppose  one  of  these  parts  is  contained  in 
a  more  than  m  times  and  less  than  m  -f  1  times.     Then  the 

ratio  ~  >  — ,  but  <  — ^t_  j    that  is,  the  value  of  -  lies 
0       n  n  0 

between  —  and  ■ — 

n  n 

The  error,  therefore,  in  taking  either  of  these  values  for 

-  is  less  than  —     But  by  increasing  n  indefinitely,  -  can 
b  n  n 

be  made  to  decrease  indefinitely,  and  to  become  less  than 

any  assigned  value,  however  small,  though  it  cannot  be 

made  absolutely  equal  to  zero. 

Hence,  the  ratio  of  two  incommensurable  quantities  can- 
not be  expressed  exactly  in  figures,  but  it  may  be  expressed 
approximately  to  any  desired  degree  of  accuracy. 

Thus,  if  h  represent  the  side  of  a  square,  and  a  the 
diagonal, 

f-V2. 


302  SCHOOL   ALGEBRA. 

Now  V2  =  1.41421356 ,  a  value  greater  than  1.414213, 

but  less  than  1.414214. 

If,  then,  a  millionth  part  of  h  is  taken  as  the  unit,  the 

value  of  the  ratio  -  lies  between  - — — — -  and  , 

b  1000000  1000000' 

and  therefore  differs  from  either  of  these  fractions  by  less 

than 

1000000 

By  carrying  the  decimal  farther,  a  fraction  may  be  found 
that  will  differ  from  the  true  value  of  the  ratio  by  less  than 
a  billionth,  a  trillionth,  or  any  other  assigned  value  what- 
ever. 

326.  The  ratio  of  two  incommensurable  quantities  is  an 
incommensurable  ratio,  and  is  a  fixed  value  toward  which 
its  successive  approximate  values  constantly  tend  as  the 
error  is  made  less  and  less. 

327.  Proportion  of  Quantities.  In  order  for  four  quanti- 
ties, A,  B,  C,  D,  to  be  in  proportion,  A  and  B  must  be  of 
the  sarae  kind,  and  C  and  D  of  the  same  kind  (but  Cand 
D  need  not  necessarily  be  of  the  same  kind  as  A  and  B), 
and  in  addition  the  ratio  of  ^  to  ^  must  be  equal  to  the 
ratio  of  C  to  D. 

If  this  is  true,  we  have  the  proportion 

A'.B=C:D. 

When  four  quantities  are  in  proportion,  the  numbers  by 
which  they  are  expressed  are  four  abstract  numbers  in 
proportion. 

328.  The  laws  of  §  316,  which  apply  to  proportion  of 
numbers,  apply  also  to  proportion  of  quantities,  except  that 
alternation  will  apply  only  when  the  four  quantities  in 
proportion  are  all  of  the  same  kind. 


RATIO    AND    PROPORTION.  303 

Exercise  105. 

1.  Find  the  duplicate  of  the  ratio  3  :  4. 

2.  Find  the  ratio  compounded  of  the  ratios  2  :  3,  3  :  4, 
6  :  7,  14  :  8. 

3.  Find  a  third  proportional  to  21  and  28. 

4.  Find  a  mean  proportional  between  6  and  24. 

5.  Find  a  fourth  proportional  to  3,  5,  and  42. 

6.  Finda;if  5  +  a7:ll-a:  =  3:5. 

7.  Find  the  number  which  must  be  added  to  both  the 
terms  of  the  ratio  3:5  in  order  that  the  resulting  ratio 
may  be  equal  to  the  ratio  15  :  16. 

li  a  :  h  =  c  :  d,  show  that 

8.  ac:bd  =  c^:d\  10.    a' —  b' :  c' -  d"  =  a" :  c\ 

9.  ab:cd=d':c\  11.    2a-\-b  :2c-{- d=b  :  d. 

12.  5a  —  b:5c  —  d=a:c. 

13.  a  —  db:a  +  Sb  =  c  —  Sd:c-{-Sd. 

14.  a''-i-ab-{-b':a''~ab  +  b'  =  c'+cd-{-d':c^-cd-{-d'. 

Find  a:  in  the  proportion 

15.  45:  68  =  90:  or.  17.    a; :  4  =  1-f- :  1-f . 

16.  6:3  =  0;:  7.  18.    3  :  a:=  7  :  42. 

19.  Find  two  numbers  in  the  ratio  2 :  3,  the  sum  of  whose 
squares  is  325. 

20.  Find  two  numbers  in  the  ratio  5  :  3,  the  difference 
of  whose  squares  is  400. 

21.  Find  three  numbers  which  are  to  each  other  as 
2:3:5,  such  that  half  the  sum  of  the  greatest  and  least 
exceeds  the  other  by  25. 


304  SCHOOL    ALGEBRA. 

22.  ¥mdxii&x  —  a:4:X—h=^^x-{-h:2x-{-a, 

23.  Find  x  and  y  from  the  proportions 

x:7/  =  x-\-i/:4:2;     x  :  y  —  x  —  i/  :  6. 

24.  Find  x  and  y  from  the  proportions 

2a;  +  y:y  =  3y:2y-a;; 
2rr+l:2a;+6  =  y:y  +  2. 

25.  If  a  +  ^  +  ^  +  ^^a-^+.-^^  ,1^^^  ^j^^^  a     c 

a-\-o  —  c  —  a     a  —  b~o-\-  a  h     a 

Variation. 

329.  A  quantity  which  in  any  particular  problem  has  a 
fixed  value  is  called  a  constant  quantity,  or  simply  a  constant ; 
a  quantity  which  may  change  its  value  is  called  a  variable 
quantity,  or  simply  a  variable. 

Variable  numbers,  like  unknown  numbers,  are  generally 
represented  by  x,  y,  z,  etc. ;  constant  numbers,  like  known 
numbers,  by  a,  S,  c,  etc. 

330.  Two  variables  may  be  so  related  that  when  a  value 
of  one  is  given,  the  corresponding  value  of  the  other  can  be 
found.  In  this  case  one  variable  is  said  to  be  a  function 
of  the  other ;  that  is,  one  variable  depends  upon  the  other 
for  its  value.  Thus,  if  the  rate  at  which  a  man  walks  is 
known,  the  distance  he  walks  can  be  found  when  the  time 
is  given  ;  the  distance  is  in  this  case  2^.  function  of  the  time. 

331.  There  is  an  unlimited  number  of  ways  in  which 
two  variables  may  be  related.  We  shall  consider  in  this 
chapter  only  a  few  of  these  ways. 

332.  When  x  and  y  are  so  related  that  their  ratio  is 
constant,  y  is  said  to  vary  as  x ;  this  is  abbreviated  thus : 


VARIATION.  305 

yccx.     The  sign  oc,  called  the  sign  of  variation,  is  read 
"varies  as."     Thus,  the  area  of  a  triangle  with  a  given 
base  varies  as  its  altitude  ;  for,  if  the  altitude  is  changed 
in  any  ratio,  the  area  will  be  changed  in  the  same  ratio. 
In  this  case,  if  we  represent  the  constant  ratio  by  m, 

y 

y  :  a:  =  m,  or  -  =  m ;  .".  y  =  mx. 

"  X 

Again,  if  y\  x'  and  3/",  a;"  be  two  sets  of  corresponding 
values  of  y  and  x,  then 

y' :  x'  =  y"  :  x", 
or  y' :  y"  =  x'  :  x".  §  316,  V. 

333.  When  x  and  y  are  so  related  that  the  ratio  of  y  to  - 
is  constant,  y  is  said  to  vary  inversely  as  x ;  this  is  written 
y  00  -.     Thus,  the  time  required  to  do  a  certain  amount  of 

X 

work  varies  inversely  as  the  number  of  workmen  employed  ; 
for,  if  the  number  of  workmen  be  doubled,  halved,  or 
changed  in  any  other  ratio,  the  time  required  will  be 
halved,  doubled,  or  changed  in  the  inverse  ratio. 

In  this  case,  y  :-  =  m ;    .'.  y  =  —,  and  a:y  =  m  ;   that  is, 
the  product  xy  is  constant. 
As  before, 


y' 

1 

=y": 

1 
V'' 

xy 

=  x'Y, 

y' 

■■y" 

=  x": 

.x'. 

or,  y' :  y"  =  a;"  :  x\  §  315 

334.   If  the  ratio  of  y :  .rz  is  constant,  then  y  is  said  to 
vary  jointly  as  x  and  z. 

In  this  case,    y  =  mxz, 

and  y' :  y"  =  a;'z' :  x"z". 


mx 

y 

y"-  ,, 

infix'' 

x' 

z' 

x" 

2" 

306  SCHOOL   ALGEBRA. 

335.    If  the  ratio  y :  -  is  constant,  then  y  varies  directly 

z 

as  X  and  inversely  as  z. 
In  this  case, 

and 


336.   Theorems. 

I.  li  y  oz  X,  and  x  cc  z,  then  y  ccz. 
For  y  =  mx  and  x  =  nz. 

.*.  y  =  mnz ; 
.*.  y  varies  as  z. 

II.  If  y  oca;,  and  ;$!oca;,  then  (y  ±2)  xa;. 
For  y  =  mx  and  z  =  nx, 

.'.  y  ±  z  =  (m±n)x] 
.'.  y±z  varies  as x. 

III.  If  ycc  X  when  z  is  constant,  and  y  oc  z  when  x  is 
constant,  then  y  cc  xz  when  a;  and  z  are  both  variable. 

Let  x',  y',  z',  and  x",  y",  z"  be  two  sets  of  corresponding 
values  of  the  variables. 

Let  X  change  from  x'  to  x",  z  remaining  constant,  and  let 
the  corresponding  value  of  y  be  Y. 

Then  y''.Y=x''.x'\  (1) 

Now  let  z  change  from  z'  to  z",  x  remaining  constant. 


Then                    r:y"  =  z':z". 

(2) 

From  (1)  and  (2), 

y'r:y"r-a;V:a;V', 

§320 

or                             y    ly"    =a;'z':a;"z", 

or                             y    :x'z'  =y''   :x"z'\ 

§  316,  V. 

y 

\  the  ratio  —  is  constant,  and  y  varies  as  xz. 
xz 


VARIATION.  307 

In  like  manner  it  may  be  shown  that  if  y  varies  as  each 
of  any  number  of  quantities  x,  z,  u,  etc.,  when  the  rest  are 
unchanged,  then  when  they  all  change,  y  oc  xzu,  etc.  Thus, 
the  area  of  a  rectangle  varies  as  the  base  when  the  altitude 
is  constant,  and  as  the  altitude  when  the  base  is  constant, 
but  as  the  product  of  the  base  and  altitude  when  both  vary. 

837.  Dxamples. 

(1)  If  y  varies  inversely  as  x,  and  when  y  =  2  the  cor- 
responding value  of  X  is  36,  find  the  corresponding  value 
of  X  when  y  =  9. 

Here  y  =  ~"i  ^^  '^  ="  ^y* 

X 

.'.  m  =  2  X  36  =  72. 
If  9  and  72  be  substituted  for  y  and  m  respectively  in 

X 

the  result  is  9  =  — ,  or  9  a;  =  72. 

X 

.*.  x  =  8.  Ans. 

(2)  The  weight  of  a  sphere  of  given  material  varies  as 
its  volume,  and  its  volume  varies  as  the  cube  of  its  diam- 
eter. If  a  sphere  4  inches  in  diameter  weighs  20  pounds, 
find  the  weight  of  a  sphere  5  inches  in  diameter. 

Let  TT  represent  the  weight, 

V  represent  the  volume, 
D  represent  the  diameter. 
Then  TFoc  F  and   Foe  D'. 

.-.  WocD^.  §336,  L 

Put  W=-  mlfi ; 

then,  since  20  and  4  are  corresponding  values  of  TTand  D, 
20  =  m  X  64. 
.    «.     20      5 

.'.  when  D  =  5,  Tr=  ^-^  of  125  =  S9^. 


308  SCHOOL   ALGEBRA. 

Exercise  106. 

1 .  li  X  ccy,  and  if  y  =  3  when  a;  =  5,  find  x  when  y  is  5. 

2.  If  TF"  varies  inversely  as  P,  and  TT  is  4  when  P  is 
15,  find  TT when  Pis  12. 

3.  li  X  ozy  and  y  ccz,  show  that  arz  oc  y*. 

4.  If  a;  oc  -  and  y  oc  -,  show  that  x  ccz. 

y  z 

5.  If  a;  varies  inversely  as  y"^ — 1,  and  is  equal  to  24 
when  y  =  10,  find  x  when  y  =  5. 

6.  If  a;  varies  as  -  +  -,  and  is  equal  to  3  when  y— 1 

y      z 

and  0  =  2,  show  that  xyz  =  2  (y  +  2). 

7.  If  :r  — y  varies  inversely  as  z  +  -,  and  x-\-y  varies 

1  .  ^ 

inversely  as  z ,  find  the  relation  between  x  and  z  if 

a;=  1,  y  =  3,  when  z 


2 


8.  The  area  of  a  circle  varies  as  the  square  of  its  radius, 
and  the  area  of  a  circle  whose  radius  is  1  foot  is  3.1416 
square  feet.  Find  the  area  of  a  circle  whose  radius  is  20 
feet. 

9.  The  volume  of  a  sphere  varies  as  the  cube  of  its 
radius,  and  the  volume  of  a  sphere  whose  radius  is  1  foot  is 
4.188  cubic  feet.  Find  the  volume  of  a  sphere  whose  radius 
is  2  feet. 

10.  If  a  sphere  of  given  material  3  inches  in  diameter 
weighs  24  lbs.,  how  much  will  a  sphere  of  the  same  material 
•weigh  if  its  diameter  is  5  inches  ? 


VABIATION.  ^     309 

11.  The  velocity  of  a  falling  body  varies  as  the  time 
during  which  it  has  fallen  from  rest.  If  the  velocity  of  a 
falling  body  at  the  end  of  2  seconds  is  64  feet,  what  is  its 
velocity  at  the  end  of  8  seconds  ? 

12.  The  distance  a  body  falls  from  rest  varies  as  the 
square  of  the  time  it  is  falling.  If  a  body  falls  through  144 
feet  in  3  seconds,  how  far  will  it  fall  in  5  seconds  ? 

The  volume  of  a  right  circular  cone  varies  jointly  as 
its  height  and  the  square  of  the  radius  of  its  base. 

If  the  volume  of  a  cone  7  feet  high  with  a  base  whose 
radius  is  3  feet  is  66  cubic  feet : 

13.  Compare  the  volume  of  two  cones,  one  of  which  is 
twice  as  high  as  the  other,  but  with  one  half  its  diameter. 

14.  Find  the  volume  of  a  cone  9  feet  high  with  a  base 
whose  radius  is  3  feet. 

15.  Find  the  volume  of  a  cone  7  feet  high  with  a  base 
whose  radius  is  4  feet. 

16.  Find  the  volume  of  a  cone  9  feet  high  with  a  base 
whose  radius  is  4  feet. 

17.  The  volume  of  a  sphere  varies  as  the  cube  of  its 
radius.  If  the  volume  is  179|-  cubic  feet  when  the  radius 
is  3-|-  feet,  find  the  volume  when  the  radius  is  1  foot  6 
inches. 

18.  Find  the  radius  of  a  sphere  whose  volume  is  the  sum 
of  the  volumes  of  two  spheres  with  radii  3-J-  feet  and  6  feet 
respectively. 

19.  The  distance  of  the  offing  at  sea  varies  as  the  square 
root  of  the  height  of  the  eye  above  the  sea-level,  and  the 
distance  is  3  miles  when  the  height  is  6  feet.  Find  the 
distance  when  the  height  is  24  feet. 


CHAPTER  XXII. 

PROGRESSIONS. 

338.  A  succession  of  numbers  that  proceed  according  to 
some  fixed  law  is  called  a  series ;  the  successive  numbers  are 
called  the  terms  of  the  series. 

A  series  that  ends  at  some  particular  term  is  a  finite  series ; 
a  series  that  continues  without  end  is  an  infinite  series. 

339.  The  number  of  different  forms  of  series  is  unlimited ; 
in  this  chapter  we  shall  consider  only  Arithmetical  Series, 
Geometrical  Series,  and  Harmonical  Series. 

Arithmetical  Progression. 

340.  A  series  is  called  an  arithmetical  series  or  an  arith- 
metical progression  when  each  succeeding  term  is  obtained 
by  adding  to  the  preceding  term  a  constant  difference. 

The  general  representative  of  such  a  series  will  be 

a,  a-{-d,  a-}- 2d,   a-f  3c? , 

in  which  a  is  the  first  term  and  d  the  common  difference ; 
the  series  will  be  increasing  or  decreasing  according  as  d  is 
positive  or  negative. 

341.  The  nth  Term.  Since  each  succeeding  term  of  the 
series  is  obtained  by  adding  d  to  the  preceding  term,  the 
coefficient  of  d  will  always  be  one  less  than  the  number  of 
the  term,  so  that  the  nth  term  is  a  +  (n  —  1)  d. 

If  the  nth  term  is  represented  by  I,  we  have 

l  =  a'i-(n-l)d.  I. 


AEITHMETICAL    PROGRESSION.  311 

342.  Sum  of  the  Series.  If  I  denotes  the  nth  term,  a  the 
first  term,  n  the  number  of  terms,  d  the  common  difference, 
and  s  the  sum  of  n  terms,  it  is  evident  that 

5=     a     +(a-{-d)i-(a+2d)-\- -}-(l -d)+     I,    or 

s=      I      ^(l-d)-^(l-2d)-{- +  (a  +  d)+     a, 

=  n(a-\-  I). 
.:    s=^(a+l).  11. 

343.  From  the  equations 

l=a  +  (n-l)d,  I. 

5  =  |(a+Z),  II. 

any  two  of  the  five  numbers  a,  d,  I,  n,  s,  may  be  found  when 
the  other  three  are  given. 

(1)  Find  the  sum  of  ten  terms  of  the  series  2,  5,  8,  11 

Here  a  =  2,  c?  =  3,  n  =  10. 

From  I.,  Z  =  2  +  27  =  29. 

Substituting  in  II.,         s  =  —  (2  +  29)  =  155.  Ans. 

A 

(2)  The  first  term  of  an  arithmetical  series  is  3,  the  last 
term  31,  and  the  sum  of  the  series  136.     Find  the  series. 

From  I.,  31-3  +  (n-l)d  (1) 

From  II.,  136  =  -(3  +  31).  (2) 

A 

From  (2),  n  =  8. 

Substituting  in  (1),         c?  =  4. 

The  series  is  3,  7,  11,  15,  19,  23,  27,  31. 


312  SCHOOL   ALGEBRA. 

(3)  How  many  terms  of  the  series  5,  9,  13, ,  must  be 

taken  in  order  that  their  sum  may  be  275  ? 

From  I.,  ?  =  5  +  (n-l)4. 

.-.  ?=4n  +  l.  (1) 

From  II.,  275  =  -(5  +  Z).  (2) 

A 
Substituting  in  (2)  the  value  of  I  found  in  (1), 

275==-(4n  +  6), 

or  2n2  +  3n=275. 

This  is  a  quadratic  with  n  for  the  unknown  number. 
Complete  the  square, 

16n2  + 0  +  9  =  2209. 
Extract  the  root,  4  n  +  3  =  ±  47. 

Therefore,  ?z  =  ll,  or  —12^. 
We  use  only  the  positive  result. 

(4)  Find  n  when  d,  I,  s  are  given. 
From  I.,  a=l  —  {n~l)  d. 
From  II..  a  =  2111^. 


Therefore,  l  —  {n  —  l)d 


n 
2s  —  In 
n 
In  —  dn^  +  dn  =  2s  —  In, 


.'.  dn^-{2l  +  d)n==-2s. 
This  is  a  quadratic  with  n  for  the  unknown  number. 
Complete  the  square, 

^d^ri"  -{)  +  {2l  +  df=^{2l  +  df  -Sds. 
Extract  the  root, 

2dn-{2l  +  d)^±  \/{2l  +  dY-8ds. 


Therefore,  „^2l+d^^(2l+ df -8. 

2d 


ARITHMETICAL    PROGRESSION.  313 

(5)    Find  the  series  in  which  the  15th  term  is  —  25  and 
41st  term  —41. 

If  a  is  the  first  term  and  d  the  common  diflference,  the  15th  term 
is  a  +  14  d,  and  the  41st  term  is  a  +  40  d. 

(1) 
(2) 


Therefore, 

a +  Ud  = -25, 

and 

a  +  40(f  =  -41. 

Subtracting, 

-26d=     16. 

.:d=-± 
13 

Substituting  : 

in  (1), 

«  =  -16tV 

Hence,  the 

series  is 

-16^%,-17,-17A, 

^  344.  The  arithmetical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
an  arithmetical  series. 

If  a  and  b  represent  two  numbers,  and  A  their  arithmet- 
ical mean,  then  a,  A,h  are  in  arithmetical  progression,  and 
by  the  definition  of  an  arithmetical  series,  §  840, 

A  —  a  =  d, 
and  b  —  A  =  d. 

.'.  A  —  a  =  b  —  A. 

.     A—Ci-i-b 


345.    Sometimes  it  is  required  to  insert  several  arithmeti- 
cal means  between  two  numbers. 

Ex.    Insert  six  arithmetical  means  between  3  and  17. 

Here  the  whole  number  of  terms  is  eight ;  3  is  the  first  term  and 
17  the  eighth. 

By  I.,  17  =  3 +  7d 

d  =  2. 

The  series  is  3,  [5,  7,  9,  11,  13,  15,]  17,  the  terms  in 
brackets  being  the  means  required. 


314  SCHOOL   ALGEBRA. 

346.  When  the  sum  of  a  number  of  terms  in  arithmetical 
progression  is  given,  it  is  convenient  to  represent  the  terms 
as  follows : 

Three  terms  by     x  —  y,x,x-{-y\ 
four  terms  by  a;  — Sy,  x—y,x-\-y,x-^Zy\ 

and  so  on. 

Ex.  The  sum  of  three  numbers  in  arithmetical  progres- 
sion is  36,  and  the  square  of  the  mean  exceeds  the  product 
of  the  two  extremes  by  49.     Find  the  numbers. 

Let  x  —  y,  X,  x  +y  represent  the  numbers. 
Then,  adding,  3  a;  =  36. 

.-.  X  =  12. 
Putting  for  x  its  value,  the  numbers  are 

12 -y,  12,  12 +  y. 
Then       (12)^  -  (12  -  y)  (12  +y)  =  the  excess. 
But  49  =  the  excess. 

Therefore,  144  -  144  +  y*  =  49. 

.'.  y  =  ±  7. 
The  numbers  are  5,  12,  19;    or  19,  12,  5. 

Exercise  107. 
l^a  +  {n-l)d',    «-2(a  +  Z)  =  ^[2a  +  (n-l)4 

Find  I  and  s,  if 

1.  a  =  7,  c?  =  4,  w  =  13.     4.    a  =  63,  c?  =  — 5,  w  =  8. 

2.  a  =  5,  £f=3,  n=12.      6.    a=|,     d  =  %       7i  =  15. 

7  7 

8.   a==i   c?=6,  ?i  =  30.     0.   a  =  3n,  c?  =  2n,    n  =  36. 


ARITHMETICAL   PROGRESSION.  316 

Find  d  and  s,  if 

7.  a  =  2,  Z=134,  n=13. 

8.  a  =  0,  1  =  200,  71  =  51. 

9.  a  =  169,  ^  =  8,  w  =  24. 

10.    a  =  ba\   Z=145a^  71  =  21. 

Insert  eight  arithmetical  means  between 

11.  13  and  76.  13.  -  and  ~ 

2  4 

12.  1  and  2.  14.  47  and  2. 

Find  a  and  s,  if 
15.   c^=7,  ^=149,  71  =  22.      16.  c?=21,  ^  =  242,  71=12. 

Find  n  and  s,  if 
17.   a=17,  ^=350,  cf=9.      18.  a  =  34,  ^=  10,  c?= -2. 

Find  d  and  n,  if 
19.   a=l|,  Z=54,  s  =  999.     20.  a  =  2,  ^=87,  5  =  801. 

Find  d  and  Z,  if 
21.    a=  10,  71  =  14,  s=  1050.    22.  a=  1,  7i  =  20,  «  =  305. 

Find  a  and  o?,  if 
23.    ^  =  21,71  =  7,5=105.       24.^=105,71=16,5  =  840. 

Find  a  and  ^,  if 
25.    71=21,  c?  =  4,  5  =  1197.    26.  n  =  25,  5  =  — 75,  c?=i. 

Find  n  and  /,  if 
27.    5  =  636,  a  =  9,  c?=8.        28.  s  =  798,  a  =  18,  c?=6. 

Find  a  and  ti,  if 
29.   5  =  623,(^=5,^=77.       30.  5  =x  1008,  c?=  4,^=88. 


316  School  algebra. 

Find  the  arithmetical  series  in  which 

31.  The  15th  term  is  25,  and  the  29th  term  46. 

How  many  terms  must  be  taken  of 

32.  The  series  -16,  -15,  -14, to  make  -100? 

33..  The  series  20,  18|,  ITJ, to  make  1621? 

34.  The  sum  of  three  numbers  in  arithmetical  progres- 
sion is  9,  and  the  sum  of  their  squares  is  29.  Find  the 
numbers. 

35.  The  sum  of  three  numbers  in  arithmetical  progres- 
sion is  12,  and  their  product  is  60.     Find  the  numbers. 

Geometrical  Progression. 

347.  A  series  is  called  a  geometrical  series  or  a  geometrical 
progression  when  each  succeeding  term  is  obtained  by  mul- 
tiplying the  preceding  term  by  a  constant  multiplier. 

The  general  representative  of  such  a  series  will  be 

a,  ar,  ar^,  ar^,  ar* , 

in  which  a  is  the  first  term  and  r  the  constant  multiplier 
or  ratio. 

The  terms  increase  or  decrease  in  numerical  magnitude 
according  as  r  is  numerically  greater  than  or  numerically 
less  than  unity. 

348.  The  nth  Term.  Since  the  exponent  of  r  increases  by 
one  for  each  succeeding  term  after  the  first,  the  exponent 
will  always  be  one  less  than  the  number  of  the  term,  so 
that  the  nth  term  is  ar'^~^. 

If  the  nth  term  is  represented  by  /,  we  have 

l^ar'^K  I. 


GEOMETRICAL    PROGRESSION.  317 

349.  Sum  of  the  Series.  If  I  represent  the  wth  term,  a  the 
first  term,  n  the  number  of  terms,  r  the  common  ratio,  and 
5  the  sum  of  n  terms,  then 

s  =  a  +  ar  -f  ar'  + ar'*"^  (1) 

Multiply  by  r, 

rs  =  ar -\-  ar^  -\- ar^  -\- ar'^~'^  +  tt^**-  (2) 

Subtracting  the  first  equation  from  the  second, 
rs  —  s=-  ar^  —  a, 
or  (r  —  1)  s  =  a  {f  ~  1). 


r-1 
Since  ^  =  af~^,  rl.=  ar'^,  and  II.  may  be  written 


II. 


s  = -.  HI. 

r—  1 


350.  From  the  two  equations  I.  and  II.,  or  the  two  equa- 
tions I.  and  III.,  any  two  of  the  five  numbers  a,  r,  I,  n,  s, 
may  be  found  when  the  other  three  are  given. 

(1)  The  first  term  of  a  geometrical  series  is  3,  the  last 
term  192,  and  the  sum  of  the  series  881.  Find  the  number 
of  terms  and  the  ratio. 

(1) 

(2) 


From  I., 

192  =  3r"-i, 

From  III., 

3S,_192r-3, 
r-1 

From  (2), 

r=2. 

Substituting 

in  (1). 

2«-i  =  64. 

The  series  is  3,  6,  12,  24,  48,  96,  192. 


318  SCHOOL    ALGEBRA. 

(2)   Find  I  when  r,  n,  s  are  given. 

From  I.,  a  =  — -• 

yfi — 1 

Substituting  in  III.,  s  = — » 

r—  1 

351.  The  geometrical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
a  geometrical  series. 

If  a  and  b  denote  two  numbers,  and  O  their  geometrical 
mean,  then  a,  O,  b  are  in  geometrical  progression,  and  by 
the  definition  of  a  geometrical  series,  §  347, 

r. 


a 

=  r,  and 

b 
G 

.    0^ 
a 

b 
~  0 

.-.  G  = 

=  Vab. 

352,    Sometimes  it  is  required  to  insert  several  geometri- 
cal means  between  two  numbers. 

Ex.    Insert  three  geometrical  means  between  3  and  48. 

Here  the  whole  number  of  terms  is  five ;  3  is  the  first  term  and  48 
the  fifth. 

By  I.,  48  =  3»^, 

r*  =  16. 
r  =  ±2. 
The  series  is  one  of  the  following : 

3,    [     6,     12,        24],    48; 

3,     [-6,     12,     -24,]    48. 

The  terms  in  brackets  are  the  means  required. 


GEOMETRICAL    PROGRESSION.  319 

353.   Infinite  Geometrical  Series.    When  r  is  less  than  1,  the 

successive  terms  become  numerically  smaller  and  smaller ; 

by  taking  n  large  enough  we  can  make  the  nth  term,  ai^"^, 

as  small  as  we  please,  although  we  cannot  make  it  absolutely 

zero. 

The  sum  of  n  terms,      ~f,  may  be  written  — -^ -^ —  ; 

r  — 1  1  — r     1— r 

this  sum  differs  from by  the  fraction ;  by  taking 

1  —  r  1  — r 

enough  terms  we  can  make  l^  and  consequently  the  fraction 

,  as  small  as  we  please ;   the  greater  the  number  of 

1  — r 

terms  taken  the  nearer  does  their   sum  approach   -.  _   • 

Hence  ^ is  called  the  sum  of  an  infinite  number  of 

1— r 

terms  of  the  series. 

(1)  Find  the  sum  of  the  infinite  series  1  — q^~7~"5~'" 

Z      4      o 

Here,  a  =  l,  r^ 

2 

1  2 

The  sum  of  the  series  is or  -.  An%. 

1  +  i        3 

We  find  for  the  sum  of  n  terms  — j —  (  — )       ;  this  sum  evidently 
2  3     3  \     2/ 

approaches  -  as  n  increases. 

(2)  Find  the  value  of  the  recurring  decimal 

0.12135135 

Consider    first    the    part    that    recurs;    this    may    be    written 

135 

135              135        ,  ,   ,,  »,,...      100000 

+ ,  and  the  sum  of  this  series  is 


100000     100000000  J 1_ 

^  1000 

which  reduces  to  — -.     Adding  0.12,  the  part  that  does  not  recur,  we 

10  1  AilO 

obtain  for  the  value  of  the  decimal  —  +  — ,  or  .  Am. 

100     740         3700 


320  SCHOOL   ALGEBRA. 


Exercise  108. 


2  =  a,n-i.,^^i!^lzil}  =  ?lll^. 
r— 1         r— 1 

Find  I  and  5,  if 

1.    a  =  4,  r  =  2,  w  =  7.         2.    a  =  9,  r  =  i   n  =  ll. 

o 

Find  r  and  s,  if 

3.  a  =  l,  n  =  4,  ^=64. 

4.  a=7,  w  =  8,  ^  =  896. 

5.  Insert  1  geometrical  mean  between  14  and  686. 

6.  Insert  3  geometrical  means  between  31  and  496. 

7.  Findaands,  if  Z=128,  r  =  2,  w=7. 

8.  Find  s  and  n,  if  a  =  9,  Z  =  2304,  r  =  2, 

9.  Find  r  and  n,  if  a  ==  2,  1=  1458,  s  =  2186. 

Q  OO 

10.  If  the  5tb  term  is  -  and  the  7th  term  — ,  find  the 

u  Ox 

series. 

11.  Find  three  numbers  in  geometrical  progression  whose 
sum  is  14,  and  the  sum  of  whose  squares  is  84. 


Sum  to  infinity : 

"•'■H^ 

14. 

2   2    2    _ 
^'  7'  49' 

.       16. 

ill. 

''  4'  16' 

"■«.2.i. 

15. 

^'^'1 

17. 

3.-2,| 

Find  the  value  of 

18.    0.16. 

20. 

0.86. 

22. 

0.736. 

19.    0.378. 

21. 

0.54. 

23. 

0.363. 

harmonical  pbogression.  321 

Harmonical  Progression. 

354.  A  series  is  called  a  hannonical  series,  or  a  harmonical 
progression,  when  the  reciprocals  of  its  terms  form  an  arith- 
metical series. 

The  general  representative  of  such  a  series  will  be 

1^       1  1  ^  1 

a     a-\-d    a -{-2d  a  +  (w  — l)c? 

Questions  relating  to  harmonical  series  are  best  solved 
by  writing  the  reciprocals  of  its  terms,  and  thus  forming 
an  arithmetical  series. 

355.  If  a  and  b  denote  two  numbers,  and  JI  their  har- 
monical mean,  then,  by  the  definition  of  a  harmonical  series, 

B:     a     b     H' 

2ab 


IT 


a-\-b 


356.    Sometimes  it  is  required  to  insert  several  harmoni- 
cal means  between  two  numbers. 


Ex.    Insert  three  harmonical  means  between  3  and  18. 

3  18 


Find  the  three  arithmetical  means  between  -  and  — -• 


These  are  found  to  be  — ,  — ,  —  ;  therefore,  the  harmonical  means 

(79  179  ^79  72  72  72 


357.    Since  ^  =^-±i,  and  G=Vab,  and  H 


2ab 


2    ' '  a +  6' 

G' 


ir=  ^,  and    G  =  ^AH. 

A 

That  is,  the  geometrical  mean  between  two  numbers  is 
also  the  geometrical  mean  between  the  arithmetical  and 
harmonical  means  of  the  numbers. 


322  SCHOOL   ALGEBRA. 


Exercise  109. 

1.  If  a,  b,  c  are  in  harmonical  progression,  show  that 
a  —  h:b  —  c==a:c. 

2.  Show  that  if  the  terms  of  a  harmonical  series  are  all 
multiplied  by  the  same  number,  the  products  will  form  a 
harmonical  progression. 

3.  The  second  term  of  a  harmonical  series  is  2,  and  the 
fourth  term  6.     Find  the  series. 

4.  Insert  the  harmonica!  mean  between  2  and  3. 

5.  Insert  2  harmonical  means  between  1  and  -• 

4 

6.  Insert  5  harmonical  means  between  1  and  -• 

7 

7.  The  first  term  of  a  harmonical  progression  is  1,  and 
the  third  term  -.     Find  the  8th  term. 

8.  The  first  term  of  a  harmonical  progression  is  1,  aijd 
the  sum  of  the  first  three  terms  is  If.     Find  the  series. 

9.  If  a  is  the  arithmetical  mean  between  b  and  c,  and  b 
the  geometrical  mean  between  a  and  c,  show  that  c  is  the 
harmonical  mean  between  a  and  b. 

10.  The  arithmetical  mean  between  two  numbers  exceeds 
the  harmonical  mean  by  1,  and  twice  the  square  of  the 
arithmetical  mean  exceeds  the  sum  of  the  squares  of  the 
harmonical  and  geometrical  means  by  11.  Find  the  num- 
bers. 


CHAPTER  XXIII. 
PROPERTIES    OF    SERIES. 

358.  Oonyergent  and  Divergent  Series.     By  performing  the 

indicated  division,  we  obtain  from  the  fraction the 

l  —  x 

infinite  series  l-f^  +  ^'  +  ^  + •     This  series,  however, 

is  not  equal  to  the  fraction  for  all  values  of  x, 

359.  If  X  is  numerically  less  than  1,  the  series  is  equal  to 
the  fraction.  In  this  case  we  can  obtain  an  approximate 
value  for  the  sum  of  the  series  by  taking  the  sum  of  a 
number  of  terms ;  the  greater  the  number  of  terms  taken, 
the  nearer  will  this  approximate  sum  approach  the  value  of 
the  fraction.  The  approximate  sum  will  never  be  exactly 
equal  to  the  fraction,  however  great  the  number  of  terms 
taken  ;  but  by  taking  enough  terms,  it  can  be  made  to  diflfer 
from  the  fraction  as  little  as  we  please. 

Thus,  if  a;  =  -,  the  value  of  the  fraction  is  2,  and  the 
series  is 

1  +  1  +  1  +  1+ 

The  sum  of  four  terms  of  this  series  is  l-J;  the  sum  of 
five  terms,  1||-;  the  sum  of  six  terms,  1-|^;  and  so  on. 
The  successive  approximate  sums  approach,  but  never 
reach,  the  finite  value  2. 

360.  An  infinite  series  is  said  to  be  convergent  when  the 
sum  of  the  terms,  as  the  number  of  terms  is  indefinitely  in- 
creased, approaches  some  fixed  finite  value;  this  finite  value 
is  called  the  sum  of  the  series. 


324  SCHOOL   ALGEBRA. 

361.  In   the   series   l-{-x-\-x^-\-af-{-  suppose  x 

numerically  greater  than  1.  In  this  case,  the  greater  the 
number  of  terms  taken,  the  greater  will  their  sum  be ;  by 
taking  enough  terms,  we  can  make  their  sum  as  large  as 
we  please.  The  fraction,  on  the  other  hand,  has  a  definite 
value.  Hence,  when  x  is  numerically  greater  than  1,  the 
series  is  not  equal  to  the  fraction. 

Thus,  if  ^  =  2,  the  value  of  the  fraction  is  —1,  and  the 
series  is 

1  +  2  +  4  +  8  + 

The  greater  the  number  of  terms  taken,  the  larger  the  sum. 
Evidently  the  fraction  and  the  series  are  not  equal. 

362.  In  the  same  series  suppose  x=l.  In  this  case  the 
fraction  is =  -,  and  the   series   1  +  1  +  1  +  1  + 

The  more  terms  we  take,  the  greater  will  the  sum  of  the 
series  be,  and  the  sum  of  the  series  does  not  approach  a 
fixed  finite  value. 

1(  X,  however,  is  not  exactly  1,  but  is  a  little  less  than  1, 

the  value  of  the  fraction ■  will  be  very  great,  and  the 

X        X 

fraction  will  be  equal  to  the  series. 

Suppose  x  =  —l.     In  this  case  the  fraction  is =  -, 

^  1  +  12 

and  the  series  1  —  1  +  1  —  1  + If  we  take  an  even 

number  of  terms,  their  sum  is  0  ;  if  an  odd  number,  their 
sum  is  1.     Hence  the  fraction  is  not  equal  to  the  series. 

363.  A  series  is  said  to  be  divergent  when  the  sum  of  the 
terms,  as  the  number  of  terms  is  indefinitely  increased, 
either  increases  without  end,  or  oscillates  in  value  without 
approaching  an^  fixed  finite  value. 


PROPERTIES   OF   SERIES.  S26 

No  reasoning  can  be  based  on  a  divergent  series ;  hence, 
in  using  an  infinite  series  it  is  necessary  to  make  such 
restrictions  as  will  cause  the  series  to  be  convergent.    Thus, 

we  can  use  the  infinite  series  1 +  a7-f-a;' +  ^  + when, 

and  only  when,  x  lies  between  -f  1  and  —1. 

364.  Identical  Series.  If  two  series,  arranged  hy  powers 
of  X,  are  equal  for  all  values  of  x  that  make  both  series  con- 
vergent, the  corresponding  coefficients  are  equal  each  to  each. 

Yov/iiA  +  Bx-\-Cx''-\- =  ^'  +  5'a;+C''^+ , 

by  transposition, 

A-A'={B'-B)x-\-{Q'--C)x'-\- 

Now,  by  taking  x  sufficiently  small,  the  right  side  of  this 
equation  can  be  made  less  than  any  assigned  value  what- 
ever, and  therefore  less  than  A  —  A\  if  A  — A'  have  any 
value  whatever.     Hence,  A  —  A^  cannot  have  any  value. 

Therefore, 

^-^'=0  or  A  =  A\ 

Hence,  £x -{- Cx' +  JDx^  + =  B'x  +  C'x^  +  B'x"  + 

or  (B-£')x  =  (0'-C)x'  +  (I)'-I))x^+ ; 

by  dividing  by  x, 

B-B  =  {C^-C)x-\-  (B'-I))x'  + ; 

and,  by  the  same  proof  as  for  A  —  A\ 
B-B'  =  0  or  B  =  B'. 
In  like  manner, 

C=  C\    D  =  D\  and  so  on. 
Hence,  the  equation 

A-\-Bx-{-Cx'-\- ^A'  +  B'x+  Cx"-}- , 

if  true  for  all  finite  values  of  x,  is  an  identical  equation; 
that  is,  the  coefficients  of  like  powers  of  x  are  the  same. 


326  SCHOOL   ALGEBRA. 

365.  Indeterminate  Coefficients. 

2  4-  3a; 
Ex.    Expand  - — — ^  in  ascending  powers  of  x. 

X  ~j~  X  ~p  X 

Assume  2  +  ^=^    =  A  +  Bx  +  Cx^  +  Dr^ ; 

1  +x  +  x^ 

then,  by  clearing  of  fractions, 

2  +  3x  =  A  +  Bx  +  Cx^  +  Dar^  + 

+  Ax  +  Bx^  +  C3?  + 

+  Ax^  +  J5a^  + 

.-.  2^Zx  =  A^-{B  +  A)x  +  {0+B+A)a?  +  {D-\-C+B)si?  + 

By  §364,       A  =  2,  5+^  =  3,  (7  +  5+^  =  0,  X> +  (7  +  5-0; 
whence  B='l,    (7=  — 3,   D  =  2  ;    and  so  on. 

.-.  ^±l£-  =  2  +  a;-3a;2  +  2a;'  + 

1+x  +  x^ 

The  series  is  of  course  equal  to  the  fraction  for  only  such  values 
of  x  as  make  the  series  convergent. 

Note.  In  employing  the  method  of  Indeterminate  Coefficients, 
the  form  of  the  given  expression  must  determine  what  powers  of  the 
variable  x  must  be  assumed.  It  is  necessary  and  sufficient  that  the 
assumed  equation,  when  simplified,  shall  have  in  the  right  member 
all  the  powers  of  x  that  are  found  in  the  left  member. 

If  any  powers  of  x  occur  in  the  right  member  that  are  not  in  the 
left  member,  the  coefficients  of  these  powers  in  the  right  member  will 
vanish,  so  that  in  this  case  the  method  still  applies ;  but  if  any  powers 
of  X  occur  in  the  left  member  that  are  not  in  the  right  member,  then 
the  coefficients  of  these  powers  of  x  must  be  put  equal  to  0  in  equating 
the  coefficients  of  like  powers  of  x ;  and  this  leads  to  absurd  results. 
Thus,  if  it  were  assumed  that 

^  +  ^^    =Ax-\-  Bx^  +  ac»  + 

l+x-Yx"^ 

there  would  be  in  the  simplified  equation  no  term  on  the  right  cor- 
responding to  2  on  the  left ;  so  that,  in  equating  the  coefficients  of 
like  powers  of  a:,  2,  which  is  2  a;",  would  have  to  be  put  equal  to  Oa;''; 
that  is,  2  =  0,  an  absurdity. 


PROPERTIES   OP   SERIES.  327 


Exercise  110. 

Expand  to  four  terms : 
1. 


1  +  2a;  1  +  x  +  x^                      l-^-x" 

1 g  5  — 2a7  g         1+x 

2  —  Sx  *  1  +  x-x^'                 '    l+x  +  x"' 

l  +  x  ^  2-Sx  g         l-8a; 


A 

1-x 

l  +  x  +  x" 

K 

6~2x 

1  +  x-x' 

A 

2-Sx 

l-2a;+3a;^ 

terms : 

19 

5 -2a: 

l  +  3a:-a;' 

Ti 

x'-x  +  l 

S  +  x 


2  +  Sx  l-2a;+3a;^  l-x-Qx' 


10.    — ^^ —  12.    — ^^ 14 


3a:-2 


2  +  x-  l-{-Sx  —  x'  x(x  —  iy 

11.    2^1^.  13.    ^-^  +  ^.  15.      ^-x+1 


S  +  x  x(x-2)  (x-lXx^-i-l) 

366.  Partial  Practions.  To  resolve  a  fraction  into  partial 
fractions  is  to  express  it  as  the  sum  of  a  number  of  frac- 
tions of  which  the  respective  denominators  are  the  factors 
of  the  denominator  of  the  given  fraction.  This  process  is 
the  reverse  of  the  process  of  adding  fractions  which  have 
different  denominators. 

Resolution  into  partial  fractions  may  be  easily  accom- 
plished by  the  use  of  indeterminate  coefficients  and  the 
theorem  of  §  364. 

In  decomposing  a  given  fraction  into  its  simplest  partial 
fractions,  it  is  important  to  determine  what  form  the  assumed 
fractions  must  have. 

Since  the  given  fraction  is  the  sum  of  the  required  par- 
tial fractions,  each  assumed  denominator  must  be  a  factor 
of  the  given  denominator ;  moreover,  all  the  factors  of  the 
given  denominator  must  be  taken  as  denominators  of  the 
assumed  fractions. 


SCHOOL   ALGEBRA. 

Since  the  required  partial  fractions  are  to  be  in  their 
simplest  form  incapable  of  further  decomposition,  the  nu- 
merator of  each  required  fraction  must  be  assumed  with 
reference  to  this  condition.  Thus,  if  the  denominator  is 
a;"  or  (x  rh  a)**,  the  assumed  fraction  must  be  of  the  form 

4  or  —^ — ;  for  if  it  had  the  form  4^±^  or  A£±A, 

or         {X  rfc  ay  X^  (X  dr  «)" 

it  could  be  decomposed  into  two  fractions,  and  the  partial 

fractions  would  not  be  in  the  simplest  form  possible. 

"When  all  the  monomial  factors,  and  all  the  binomial 

factors,  of  the  form  x±a,  have  been  removed   from  the 

denominator  of  the  given  expression,  there  may  remam 

quadratic  factors  which  cannot  be  further  resolved;  and 

the  numerators  corresponding   to  these  quadratic  factors 

may  each  contain  the  first  power  of  ar,  so  that  the  assumed 

Ax  -f-  S 
fractions  must  have  either  the  form  5-—^- -,  or  the 

,         Ax  +  B  x'±ax  +  b 

term  — -— ; — 
x^  +  h 


3      . 

(1)   Resolve  — — -  into  partial  fractions. 

X   'Y'  •*• 


Since  o?  -\-\=-{x-\-\){x^  —  x-\-\),  the  denominators  will  be  rc  +  l 
and  x^  —  x-\-\. 

A                          3            A     ,    Bx  +  C   . 
Assume  -- — -  = + . 

ar  +  1      X  -\-l      x^  —  X  ■\-\ 

then  3  =  ^(a;2  -  a;  +  1)  +  (5a;  +  C)(a;  +  1) 

^{A-\-B)x'  +  {B-vC~A)x  +  {A^C)', 
whence,  3  =  ^  +  C,  B+C-A  ='0,  A  +  B  =  0, 

and  '     A  =  \   ^=.-1,   C7=2. 


Therefore, 


3      ^     1     _     x-2 
jp'+l     a;  +  l     a;2-a;  +  l 


f>ROPERTIES   OP   SERIES.  329 

(2)    Resolve into  partial  fractions. 

The  denominators  may  be  x,  a;',  x  +  1,  {x  +  ly. 
Assume    4x3-a^--3.- 2^  X  ^  P  ^  _CL  ^      D 


x^{x+lf  X      x^     x  +  l      {x  +  iy 

.-.  4x^-x^-Sx~2  =  Ax{x  +  lf  +  B{x  +  iy  +  Cx^x  +  -[)  +  Dx* 
=={A  +  C)a^  +  {2A  +  B+0+D)x^  +  {A  +  2B)x  +  Bi 
whence,  J.  +  (7=  4, 

2A  +  B  +  C-^D=^-1, 
A  +  2B  =  -3, 
B  =  -2- 
and  .'.  B  =  -2,     A  =  l,     C=  3,     i)=--4. 

Therefore.  ^^^-^^-^^-^^1-1 +-g ^ 

x\x  +  iy        X    x^    x  +  l    (x  +  l)* 

Exercise  111. 
Resolve  into  partial  fractions : 

7^7+1  ^      3a;»  — 4 


2. 


(a;  +  4)(a;-5)  ^7^(^  +  5) 

7:r  — 1  «  7^^  — a; 


(1  -  2x)  (1-Sx)  {x-iy\x+2) 

bx—1  g     2a;'-7a7+l 


(2ar-l)(a:-5)  ar'-l 

4.  ^-^ 10,  ^^-1 


(a:-5)(a;  +  2)  (6a;-f  l)(a: -1) 

3  11  ^'  ~  ^ 


ar'-l  (ar'+l)(:r  +  2) 

a?'  —  a;  —  3  -«  a;'  —  a:  +  1 

ar(ar^-4)*  '    (^  +  1)  (a?  -  l/ 


CHAPTER  XXIV. 

BINOMIAL    THEOREM. 

367.  Binomial  Theorem,  Positive  Integral  Exponent.  By  suc- 
cessive multiplication  we  obtain  the  following  identities : 

(a  +  hf  =  a'  +  Sa'b  +  ^ab^  +  b^; 

(a  +  by  =  a*  +  4a'^>  +  ^a^b''  +  4a6'  +  b\ 

The  expressions  on  the  right  may  be  written  in  a  form 
better  adapted  to  show  the  law  of  their  formation  : 

{a  Jr  bf  =  a? +  2ab  +p|*^ 

(«  +  if  =  a'  +  4a'J  +i;|a'6'  +^^^aV  +ff|^4'. 

Note.  The  dot  between  the  Arabic  figures  means  the  same  as  the 
sign  X. 

368.  Let  n  represent  the  exponent  of  {a  +  b)  in  any  one 
of  these  identities ;  then,  in  the  expressions  on  the  right, 
we  observe  that  the  following  laws  hold  true : 

I.    The  number  of  terms  is  n  +  I'- 
ll.   The  first  term  is  a",  and  the  exponent  of  a  is  one 

less  in  each  succeeding  term. 

The  first  power  of  b   occurs   in  the  second    term,  the 

second  power  in  the  third  term,  and  the  exponent  of  6  is 

one  greater  in  each  succeeding  term. 

The  sum  of  the  exponents  of  a  and  b  in  any  term  is  n. 


BINOMIAL   THEOEEM.  331 

III.  The  coefficient  of  the  first  term  is  1 ;  of  the  second 
term,  n ;  of  the  third  term,         '""    ^  ;   and  so  on. 

369.  Consider  the  coefficient  of  any  term  ;  the  number 
of  factors  in  the  numerator  is  the  same  as  the  number  of 
factors  in  the  denominator,  and  the  number  of  factors  in 
each  is  the  same  as  the  exponent  of  h  in  that  term  ;  this 
exponent  is  one  less  than  the  number  of  the  term. 

370.  Proof  of  the  Theorem.  That  the  laws  of  §  368  hold 
true  when  the  exponent  is  any  positive  integer,  is  shown 
as  follows : 

We  know  that   the   laws   hold  for  the  fourth  power; 
suppose,  for  the  moment,  that  they  hold  for  the  ^th  power. 
We  shall  then  have 

(a  +  hy  =  a*  +  ^oJ^-'h  +  Miizl)  a^-^b^ 
^  ^  .  1-2 

,  lc{lc—l){h-2)  ^u-zp  ,  ,jv 

1-2-3  ^^ 

Multiply  both  members  of  (1)  by  a  -f  5  ;  the  result  is 

(a  +  by+\=  a*+^  +  (k+l)  a'b  +  (^+^)^  a'-'b' 

i. '  A 

-|_(i±IlA(llllla*-*53_^ (2) 

±'  A'  o 

In  (1)  put  ^  4"  1  for  ^  ;  this  gives 
(a  +  6)*+^  =  a*+^  +  ik  +  l)  a'b  +  {^  +  '^)(^  +  '^-'^)  a'-^b' 

(^+l)(^+l-l)a+'l-2)    .-..,  , 

^  "    •     1-2-3  a     o  -f 

=  a*+^  +  (Jc  +  1)  a*i  +  ^^p|^  oJ^'h* 

^(.•+l)^.g-l)^^,,,^ (3) 

Equation  (3)  is  seen  to  be  the  same  as  equation  (2). 


332  SCHOOL   ALGEBRA. 

Hence  (1)  holds  when  we  put  ^-f  1  for  ^ ;  that  is,  if  the 
laws  of  §  368  hold  for  the  ^th  power,  they  must  hold  for 
the  {k  +  l)th  power. 

But  the  laws  hold  for  the  fourth  power ;  therefore  they 
must  hold  for  the  fifth  power. 

Holding  for  the  fifth  power,  they  must  hold  for  the  sixth 
power  ;  and  so  on  for  any  positive  integral  power. 

Therefore  they  must  hold  for  the  nth  power,  if  n  is  a 
positive  integer ;  and  we  have 


(a  +  hy  -  a"  +  na^-'b  +  ^^"""^^  a"-^5^ 


+  "("-^)^"-^)a"-W  + (A) 

L'  A'  o 

Note.   The  above  proof  is  an  example  of  a  proof  by  mathematical 
induction. 


371.  This  formula  is  known  as  the  binomial  theorem. 
The  expression  on  the  right  is  known  as  the  expansion  of 

(a  +  ^)";  this  expansion  is  definite  series  when  n  is  a  positive 
integer.     That  the  series  is  finite  may  be  seen  as  follows  : 

In  writing  out  the  successive  coefficients  we  shall  finally 
arrive  at  a  coefficient  which  contains  the  factor  n  —  n\  the 
corresponding  term  will  vanish.  The  coefficients  of  all  the 
succeeding  terms  likewise  contain  the  factor  n  —  n,  and 
therefore  all  these  terms  will  vanish, 

372.  If  a  and  h  are  interchanged,  the  identity  (A)  may 
be  written 

(a  +  by  =  (5  +  af  =  S**  +  nb^-^a  +  ^^!^~^)  b^-'a' 

^_  n{n-V){n--2)  ^„_3^3  _^ 

1.'  A'  o 


BINOMIAL   THEOREM.  333 

This  last  expansion  is  the  expansion  of  (A)  written  in 
reverse  order.  Comparing  the  two  expansions,  we  see 
that :  the  coefficient  of  the  last  term  is  the  same  as  the  co- 
efficient of  the  first  term ;  the  coefficient  of  the  last  term 
but  one  is  the  same  as  the  coefficient  of  the  first  term  but 
one ;  and  so  on. 

In  general,  the  coefficient  of  the  rth  term  from  the  end 
is  the  same  as  the  coefficient  of  the  rth  term  from  the 
beginning.  In  writing  out  an  expansion  by  the  binomial 
theorem,  after  arriving  at  the  middle  term,  we  can  shorten 
the  work  by  observing  that  the  remaining  coefficients  are 
those  already  found,  taken  in  reverse  order. 

373.  If  b  is  negative,  the  terms  which  involve  even 
powers  of  h  will  be  positive,  and  those  which  involve  odd 
powers  of  h  negative.     Hence, 

(a  -  bY  =  a^-  ndr-^b  +  ^"^-^^  oJ'-^b^ 
^  ^  1-2 

_  7i(n-l)(n-2)        3^3   ,    r^^ 

Also,  putting  1  for  a  and  a;  for  5,  in  (A)  and  (B), 

1.'  A 

^«(«-Ji(|-2)^,^ (0) 

_ni«^ll^^^ (D) 


834  SCHOOL    ALGEBRA. 

374.  Examples: 

(1)  Expand  (1  +  ^xf. 

In  (0)  put  2  a;  for  a;  and  5  for  n.     The  result  is 
(1  +  2x)^  =  1  +  5(20;)  +  |f|  4x2  +  1^  8a:» 

l'2-3-4  1-2-3-4-5 

-  1  +  10a;  +  40  a;'  +  SOar*  +  80  x*  +  32a;6. 

(2)  Expand  to  three  terms  ( J  • 

Put  a  for  -,  and  h  for  -^ ;  then,  by  (B), 
X  3 

(a  -  hf  =  a«  -  6a56  +  15  a*^'  _ 

Keplacing  a  and  h  by  their  values, 

_J__£     20_ 
a;«      x3       3 


375.  Any  Esquired  Term.  From  (A)  it  is  evident  (§  372) 
that  the  (r+  l)th  term  in  the  expansion  of  {a-{-hy  is 

9?  (ti  —  1)  {n  —  2) to  r  factors   n-rj.r 

1x2x3 r 

Note.  In  finding  the  coefficient  of  the  (r  +  l)th  term,  write 
tlie  series  of  factors  1x2x3 r  for  the  denominator  of  the  co- 
efficient, then  write  over  this  series  the  factors  n{n  —  l}(n  — 2),  etc., 
writing  just  as  many  factors  in  the  numerator  as  there  are  in  the 
denominator. 

The  (r  +  l)th  term  in  the  expansion  of  (a  — by  is  the 
same  as  the  above  if  r  is  even,  and  the  negative  of  the 
above  if  r  is  odd. 


BINOMIAL    THEOREM.  335 


\  Ex.   Find  the  eighth  term  of  ("4  - -T. 

N/  2 

Here  a  =  4,  h^-,   n=10,   r  =  \ 

2 

The  term  required  is    10-9-8-7-6-5-4(^^3^_^\ 


which  reduces  to  —  60  x^*. 

376.    A  trinomial   may  be   expanded   by  the   binomial 
theorem  as  follows: 

Expand  (1  + 2 a; -a;^. 

Put  2x-X^  =  Z; 

then  (l+z)3  =  l +32  +  322+^3. 

Replace  z  with  2x  —  x^. 

:.  (1  +  2x-x'f=l  +  3(2a;-a;2)  +  3(2a;  ^a;^  +  (2a;-a!2)5 
=  1  +  6a;  +  9a;2  -  4a:3  _  9a,4  ^  6a;5  _  a;6. 


Exercise  112. 
Expand : 

1.  (a +  5)'.  6.  (a»  +  J)'.  11.  (TO-i  +  «T 

2.  (^-2)«.  7.  (mH«T  12.  (^-+^«)«. 

^            ^'  8.  (a-iT  13.  (Sx'  +  yl)'. 

*•    (J-Ij-  9.  (a*  +  4»)'.  14.  (a*-c*/. 

6.    (4  +  3j/)'.  10.  (a-'+ *-')'■  15.  (2a'-|Va)' 


"•    ^  +  iiVa    •  20. 


/V«     1    jiV 
V^'      2Ma;' 

18.   (2a:'y-'-yVy)*.  21.   (2  aJ"' -  ia*)'. 


336  SCHOOL   ALGEBRA. 


22. 


m-y-   -  u-^j 


24.  Find  the  fourth  term  of  (2a;  —  Sy/. 

25.  Find  the  ninety-seventh  term  of  (2  a  —  bf^. 

Note.   As  the  expansion  has  101  terms,  the  ninety-seventh  term 
from  the  beginning  is  the  fifth  term  from  the  end. 

26.  Find  the  eighth  term  of  {Sx  —  i/fK 

27.  Find  the  tenth  term  of  (2  a'  -  J  a)'". 

28.  Find  the  fifth  term  of  (a  -  2Vby\ 

29.  Find  the  eleventh  term  of  (2  —  a)^^ 

30.  Find  the  fifteenth  term  of  (x  +  y)'". 

31.  Find  the  fourth  term  of  (3  -  2x''y. 

32.  Find  the  twelfth  term  of  (a'  —  aVxy\ 

33.  Find  the  seventh  term  of  (y'  —  1)^®. 
,34.  Find  the  fifth  term  of  (^a-Z»V6)^ 

35.  Find  the  fourth  term  of  (Va-V'F^^ 

36.  Find  the  third  term  of  ( Va  -  V^^)^ 

37.  Find  the  sixth  term  of  (^/a^'  —  V^)'. 

38.  Find  the  eighth  term  of  ( Vfa  +  Vp)'^ 

39.  Find  the  ninth  term  of  (:rV^  +  yV^y\ 

40.  Find  the  fifth  term  of  (a^b  -  ^AJV. 

41.  Find  the  seventh  term  of  (x  -\-  a?~^)**. 


BINOMIAL   THEOREM.  337 

377.  Binomial  Theorem,  Any  Exponent.  We  have  seen 
(§  370)  that  when  n  is  a,  positive  integer  we  have  the 
identity 

(l+x)"=l+«^+?ii^^  +  2i2i^|(p2):^+ 

We  proceed  to  the  case  of  fractional  and  negative  expo- 
nents. 

I.  Suppose  w  is  a  positive  fraction,  £..  We  may  assume 
that  ^ 

(l  +  xy=(A-{-£x-{-Cx'  +  Dx'  + y,  (1) 

provided  x  be  so  taken  that  the  series 
A  +  Bx+Cx'  +  Dx'^i- 

is  convergent  (§  360). 

That  this  assumption  is  allowable  may  be  seen  as  follows : 
Expand  both  members  of  (1).     We  obtain 

and       A'i-qA<'-'JBx  +  [^^^^^  A'^-'B'+qA'^-'CJx'+ 

In  the  first  Jc  coefficients  of  the  second  series  there  enter 

only  the  first  Ic  of  the  coefficients  A,  £,  C,  D,  If,  then, 

we  equate  the  coefficients  of  corresponding  terms  in  the 
two  series  (§  364)  as  far  as  the  ^th  term,  we  shall  have  just 

k  equations  to  find  k  unknown  numbers  A,  B,  C,  D 

Hence  the  assumption  made  in  (1)  is  allowable. 

Comparing  the  two  first  terms  and  the  two  second  terms, 
we  obtain 

^'-1,     .-.   A  =  l', 

qA'-'JB--=p,  or  qB^p,     .'.  B  =  -- 


338  SCHOOL   ALaEBRA. 

Extracting  the  ^th  root  of  both  members  of  (1),  we  have 
(1  +  a;>-  =  1  + 1  a;  +  Cx'  +  i)a;»  + ,  (2) 

where  x  is  to  be  so  taken  that  the  series  on  the  right  is 
convergent. 

II.    Suppose  n  is  a  negative  number,  integral  or  frac- 
tional.    Let  71-=  — m,  so  that  m  is  positive  ;  then 

From  (2),  whether  m  is  integral  or  fractional,  we  may 
assume 

1 1 

(l  +  aj)"*      \ -{■  mx -\- cx^ -[- dx^  ■{- 

By  actual  division  this  gives  an  equation  in  the  form 

(1  +  a:)-"*  -=l-mx+Cx''-\-  Dx^  + (3) 

378.    It  appears  from  (2)  and  (3)  §  377  th-at  whether  n  be 
integral  or  fractional,  positive  or  negative,  we  may  assume 

(l-\-xY  =  l-\-nx-{-  Cx^  +  Dx^  -f , 

provided  the  series  on  the  right  is  convergent. 
Squaring  both  members, 

(l  +  2a7  +  a;^)«=l  +  2w;r  +  2ar'4-2i)2:'  + (1) 

+  nV   +2nar'+ 

Also,  since 

(l+y)'*  =  l  +  ny+Q/'  +  %'+ , 

we  have,  putting  1x-\-x^  for  y,  * 

{jL^2x-\-xy=\-\-n{2x-^x^)-\-C{^x-\-3?)^ 

-\-D{2x-\-a?J 

=  l  +  2w2;  +  na:'    +46^*  + 

+  4ar^4-8i)a;'+ (2) 


BINOMIAL    THEOREM.  339 

Comparing  corresponding  coefficients  in  (1)  and  (2), 

n  +  4  (7=  2  (7+  n\ 
4C+8i)  =  2i)  +  2nC: 

...  20=n'-n,  and       (7=:<^-=^l 

3Z)==(«-2)C,and  J  =  "("-;^)^"-^>; 

L    A   o 
and  so  on. 

Hence,  whether  n  be  integral  or  fractional,  positive  or 
negative,  we  have 

provided,  always,  x  be  so  taken  that  the  series  on  the  right 
is  convergent. 

The  series  obtained  will  be  an  infinite  series  unless  n  is  a 
positive  integer  (§  371). 

379.   If  X  is  negative. 
Also,  if  a:  <  a, 

L  a  1  •  2      a'^  J 

=  a^  +  72a"-^x  +  ^^^~^)a"-V  + ; 

1-2  ' 

if  a;  >  a, 

(a+a;)*  =  {x"^  a)"  -  ar"  ^1  +  ^Y 

„ fi    ,     a  ,  n(n  —  V)a^  .        n 

=-"[i+«-+^iV^+ ] 

=  a^  +  woa;"-'  +  "("~-^^  aV^'  + 

1  '  Jt 


'340  SCHOOL   ALGEBRA. 

380.  Examples. 

(1)    Expand  (1  +  x)K 


3-6-9 

The  above  equation  is  only  true  for  those  values  of  x  which  make 
the  series  convergent. 

(2)  Expand    ^    ^      - 

Vl  —X 

=  "(!-«)-* 
Vl-a; 

if  X  is  so  taken  that  the  series  is  convergent. 

A  root  may  often  be  extracted  by  means  of- an  expansion. 

(3)  Extract  the  cube  root  of  344  to  six  decimal  places. 

•••^-  7(1. ^y. 

L       3V343/  1-2      V343y  J 

=  7(1  +  0.000971815  -  0.000000944+ ), 

=  7.006796. 

(4)  Find  the  eighth  term  of  fx--^-^'^- 
Here  a  =  x,     h  = =  - — -.     n  = ,     r  =  7. 


4Vx     4a;'  2 


The  term  is     — 2_! — kl — ll — L — zl — xJ — ~J. 
l-2-3'4-5-6-7 


^j,  1-3-5-7-911-13-37 


2-4-6-8-10-12-14-47-a;" 


BINOMIAL    THEOREM.  341 


Exercise  113. 

Expand  to  four  terms  . 

1.    (l+^r)^. 

6.    (l  +  x)i 

11. 

(2^+3y)t 

2.    (l  +  xy 

7.  (i  +  x)-i 

12. 

(2a;+3y)-t 

3.  (i+xy 

4.    (l-^)f. 

K     8.  a+xr\ 

9.    (l  +  bx)-'. 
I       10.    (l  +  5a;)f 

13. 
14. 

1 

</a^-x 
1 

5.    (1-^)- 

</{a-xy 

15. 

Find  the  fourth  term  of 

(- 

2VxJ 

16. 

Find  the  fifth  term  of  - 

1 

V(a  -  2xy 

17.  Find  the  third  term  of  (4 -7 a:)'. 

18.  Find  the  sixth  term  of  (a"  ~  2ax)\ 

19.  Find  the  fifth  term  of  (l-2a:)~i 

20.  Find  the  fifth  term  of  (1  -  x)-\ 

21.  Find  the  seventh  term  of  (1  —  x)^. 

22.  Find  the  third  term  of  (1  +  x)   2n. 

23.  Find  the  fourth  term  of  (1  +  x)~\ 

24.  Find  the  sixth  term  of  (^2 j      • 

25.  Find  the  fifth  term  of  (2  a:  -  3y)~^. 

26.  Find  the  fourth  term  of  (1  -  6x)~i 


CHAPTER  XXV. 
LOGARITHMS. 

381.  If  the  natural  numbers  are  regarded  as  powers  of 
ten,  the  exponents  of  the  powers  are  the  Oommon  or  Briggs 
Logarithms  of  the  numbers.  If  A  and  B  denote  natural 
numbers,  a  and  b  their  logarithms,  then  10*  =  ^,  W  =  B ; 
or,  written  in  logarithmic  form,  log^  =  a,  log  B  =  b. 

382.  The  logarithm  of  a  product  is  found  by  adding  the 
logarithms  of  its  factors. 

For  AxB=  10"  X  10*  =  10«+*. 

Therefore,        log  (^  X  ^)  -=  a  +  5  =  log  ^  +  log  B. 

383.  The  logarithm  of  a  quotient  is  found  by  subtracting 
the  logarithm  of  the  divisor  from  that  of  the  dividend. 

For  ^    :4  =  15!=:lo«-^ 

B     10* 

Therefore,       log  —  =  a  —  Z>  =  log  ^  —  log  B. 
B 

384.  The  logarithm  of  a  power  of  a  number  is  found  by 
multiplying  the  logarithm  of  the  number  by  the  exponent 
of  the  power. 

For  .         ^"  =  (10*)'*  =  10**". 

Therefore,        log^"  =  an  =  7i  log^. 


LOGARITHMS.  343 

385.  The  logarithm  of  the  root  of  a  number  is  found  by- 
dividing  the  logarithm  of  the  number  by  the  index  of  the 
root. 

For  ^/A  =  -y/W=10k 


Therefore,  log  -VA  =  -  =  — — 


386.  The  logarithms  of  1,  10,  100,  etc.,  and  of  0.1,  0.01, 
0.001,  etc.,  are  integral  numbers.  The  logarithms  of  all 
other  numbers  are  fractions. 

Since      10"=      1,  10-1  (-yV)     ^q.I, 

10^=    10,  10-'(=Ti^)   =0.01, 

10^=100,  10-H=-n^)  =  0.001. 

therefore    log     1  =  0,  log  0.1      =  —  1, 

log   10  =  1,  log  0.01    =-2, 

log  100  =  2,  log  0.001  =-3. 

Also,  it  is  evident  that  the  common  logarithms  of  all 
numbers  between 

1  and       10  will  be      0  +  a  fraction, 

10  and     100  will  be      1  +  a  fraction, 

100  and   1000  will  be      2  +  a  fraction, 

1  and  0.1  will  be  —  1  +  a  fraction, 
0.1  and  0.01  will  be  —  2  +  a  fraction, 
0.01  and  0.001  will  be  -  3  +  a  fraction. 

387.  If  the  number  is  less  than  1,  the  logarithm  is  nega- 
tive (§  386),  but  is  written  in  such  a  form  that  the  fractional 
part  is  always  positive. 

388.  Every  logarithm,  therefore,  consists  of  two  parts :  a 
positive  or  negative  integral  number,  which  is  called  the 
characteristic,  and  a  positive  proper  fraction,  which  is  called 


344  SCHOOL    ALGEBRA. 

the  mantissa.  Thus,  in  the  logarithm  3.5218,  the  integral 
number  3  is  the  characteristic,  and  the  fraction  .5218  the 
mantissa.  In  the  logarithm  0.7825  —  2,  which  is  sometimes 
written  2.7825,  the  integral  number  — 2  is  the  character- 
istic, and  the  fraction  .7825  is  the  mantissa. 

389.  If  the  logarithm  has  a  negative  characteristic,  it  is 
customary  to  change  its  form  by  adding  10,  or  a  multiple 
of  10,  to  the  characteristic,  and  then  indicating  the  sub- 
traction of  the  same  number  from  the  result.  Thus,  the 
logarithm  2.7825  is  changed  to  8.7825  —  10  by  adding  10  to 
the  characteristic  and  writing  — 10  after  the  result.  The 
logarithm  13.9278  is  changed  to  7.9273-20  by  adding  20 
to  the  characteristic  and  writing  —  20  after  the  result. 

390.  The  following  rules  are  derived  from  §  386 : 

KuLE  1.  If  the  number  is  greater  than  1,  make  the 
characteristic  of  the  logarithm  one  unit  less  than  the  num- 
ber of  figures  on  the  left  of  the  decimal  point. 

EuLE  2.  If  the  number  is  less  than  1,  make  the  charac- 
teristic of  the  logarithm  negative,  and  one  unit  more  than 
the  number  of  zeros  between  the  decimal  point  and  the  first 
significant  figure  of  the  given  number. 

Rule  3.  If  the  characteristic  of  a  given  logarithm  is 
positive,  make  the  number  of  figures  in  the  integral  part  of 
the  corresponding  number  one  more  than  the  number  of 
units  in  the  characteristic. 

Rule  4.  If  the  characteristic  is  negative,  make  the  num- 
ber of  zeros  between  the  decimal  point  and  the  first  sig- 
nificant figure  of  the  corresponding  number  one  less  than 
the  number  of  units  in  the  characteristic. 

Thus,  the  characteristic  of  log  7849.27  is  3 ;  the  character- 
istic of  log  0.037  is  —  2  =  8.0000  — 10.     If  the  characteristic 


LOGARITHMS.  -  345 

is  4,  the  corresponding  number  has  five  figures  in  its  integral 
part.  If  the  characteristic  is  —  3,  that  is,  7*0000  —  10,  the 
corresponding  fraction  has  two  zeros  between  the  decimal 
point  and  the  first  significant  figure. 

391.  The  mantissa  of  the  common  logarithm  of  any  inte- 
gral number,  or  decimal  fraction,  depends  only  upon  the 
digits  of  the  number,  and  is  unchanged  so  long  as  the 
sequence  of  the  digits  remains  the  same. 

For  changing  the  position  of  the  decimal  point  in  a 
number  is  equivalent  to  multiplying  or  dividing  the  num- 
ber by  a  power  of  10.  Its  common  logarithm,  therefore, 
will  be  increased  or  diminished  by  the  exponent  of  that 
power  of  10  ;  and  since  this  exponent  is  integral,  the  nfian- 
tissa,  or  decimal  part  of  the  logarithm,  will  be  unaffected. 

Thus,    27196  =  10*•"*^  2.7196  =  10°'^^ 

2719.6  =  lO'-*"*',         0.27196  =  10«-"^^-^«, 
27.196  =  lO'-*"'',     0.0027196  =  lO^-««-i». 

One  advantage  of  using  the  number  ten  as  the  base  of  a 
system  of  logarithms  consists  in  the  fact  that  the  mantissa 
depends  only  on  the  sequence  of  digits,  and  the  characteristic 
on  i\iQ  position  of  the  decimal  point. 

392.  In  simplifying  the  logarithm  of  a  root  the  equal 
positive  and  negative  numbers  to  be  added  to  the  logarithm 
should  be  such  that  the  resulting  negative  number,  when 
divided  by  the  index  of  the  root,  gives  a  quotient  of  — 10. 

Thus,  if  the  log  0.002^  =  i  of  (7.3010  —  10),  the  expres- 
sion J-  of  (7.3010  —  10)  may  be  put  in  the  form  \  of 
(27.3010-30),  which  is  9.1003-10,  since  the  addition  of 
20  to  the  7,  and  of  —  20  to  the  — 10,  produces  no  change 
in  the  value  of  the  logarithm. 


346  SCHOOL   ALGEBRA. 

Exercise  114. 

Given:  log 2  =  0.3010;  log 3  =  0.4771;  log 5  =  0.6990; 
log  7  =  0.8451. 

Find  the  common  logarithms  of  the  following  numbers 
by  resolving  the  numbers  into  factors,  and  taking  the  sum 
of  the  logarithms  of  the  factors : 

1.  log  6.       5.    log  25.        9.    log  0.021.      13.    log  2.1. 

2.  log  15.     6.    log  30.      10.    log  0.35.        14.    log  16. 

3.  log  21.     7.    log  42.      11.    log  0.0035.    15.    log  0.056. 

4.  log  14.     8.    log  420.    12.    log  0.004.      16.    log  0.63. 
Find  the  common  logarithms  of  the  following : 

17.  2\         20.    5^  23.    5^  26.    7'.  29.    5^ 

18.  b\         21.    2i         24.    7TT  27.    b\  30.    2^A 

19.  7*.         22.    5i         25.    2^  28.    3^^.         31.    5^ 

393.    The  logarithm  of  the  reciprocal  of  a  number  is 
called  the  cologarithm  of  the  number. 
If  A  denote  any  number,  then 

colog^  =  logi  =  logl  -  log^  (§  383)  =  -  log^  (§  386). 

Hence,  the  cologarithm  of  a  number  is  equal  to  the  log- 
arithm of  the  number  with  the  minus  sign  prefixed,  which 
sign  affects  the  entire  logarithm,  both  characteristic  and 
mantissa. 

In  order  to  avoid  a  negative  mantissa  in  the  cologarithm, 
it  is  customary  to  substitute  for  —  log^  its  equivalent 
(10  -  log  ^)- 10. 

Hence,  the  cologarithm  of  a  number  is  found  by  subtract- 
ing the  logarithm  of  the  number  from  10,  and  then  annexing 
—  10  to  the  remainder. 


LOGARITHMS.  347 

The  best  way  to  perform  the  subtraction  is  to  begin  on 
the  left  and  subtract  each  figure  of  log  J.  from  9  until  we 
reach  the  last  significant  figure,  which  must  be  subtracted 
from  10. 

If  log^  is  greater  in  absolute  value  than  10  and  less 
than  20,  then  in  order  to  avoid  a  negative  mantissa,  it  is 
necessary  to  write  —  log^  in  the  form  (20  —  log^)  —  20. 
So  that,  in  this  case,  colog  A  is  found  by  subtracting  log  A 
from  20,  and  then  annexing  —20  to  the  remainder. 

*  (1)   Find  the  cologarithm  of  4007. 

10         -10 
Given:  log  4007  =    3.6028 

Therefore,  colog  4007  =    6.3972-10 

(2)  Find  the  cologarithm  of  103992000000. 

20         -20 
Given  :  log  103992000000  =  11.0170 

Therefore,         colog  103992000000  =    8.9830  -  20 

If  the  characteristic  of  log  A  is  negative,  then  the  subtra- 
hend, —  10  or  —20,  will  vanish  in  finding  the  value  of 
colog  ^. 

(3)  Find  the  cologarithm  of  0.004007. 

10         - 10 
Given:  log  0.004007  =    7.6028-10 

Therefore,  colog  0.004007  =    2.3972 

By  using  cologarithms  the  inconvenience  of  subtracting  the 
logarithm  of  a  divisor  is  avoided.  For  dividing  by  a  num- 
ber is  equivalent  to  multiplying  by  its  reciprocal.  Hence, 
instead  of  subtracting  the  logarithm  of  a  divisor,  its  colog- 
arithm may  be  added. 


S4S  SCHOOL   ALGEBRA. 

5 


(4)    Find  the  logarithm  of 


0.002 


Iog^  =  log5  +  colog0.002. 

log  5  =  0.6990 

colog  0.002  =  2.6990 

log  quotient  -  3.3980 

(5)    Find  the  logarithm  of  -^y- 

log  0.07  =  8.8451 -10 
colog  23  =  (10  -  3  log  2)  -  10  =  9.0970  -  10 


log  quotient  =  7.9421  -  10 


Exercise  115. 


Given:  log  2  =  0.3010;  log  3  =  0.4771;  log  5  =  0.6990; 
log  7  =  0.8451 ;  log  11  =  1.0414. 

Find  the  logarithms  of  the  following  quotients  : 


1.    - 


2. 


4. 


6.    ^ 


2 
7' 

8. 

5 
ll' 

15. 

0.03 
44 

22. 

8^ 
3^' 

4 
5* 

9. 

7 
ll' 

16. 

44 
0.05 

23. 

55' 

28^' 

5 

6" 

10. 

3 

8' 

17. 

35 
0.11 

24. 

7Y2 
15^ 

7 
5* 

11. 

8 
ll' 

18. 

24 
55' 

25. 

33^ 

0.005  . 

7 
11 

12. 

5 

22' 

19. 

IP 
12^ 

26. 

0.007 
15' 

11 

7* 

13. 

0.7 
11 

20. 

5« 
6*' 

27. 

0.011 
0.05* 

5 

i 

14. 

6.05 
4 

21. 

7' 
2=^' 

28. 

0.05^ 
O.OIP 

LOGARITHMS.  349 

894.  Tables.  A  table  oi  four-place  common  logarithms 
is  given  at  the  end  of  this  chapter,  which  contains  the  com- 
mon logarithms  of  all  numbers  under  1000,  the  decimal  point 
and  characteristic  being  omitted.  The  logarithms  of  single 
digits,  1,  8,  etc.,  will  be  found  at  10,  80,  etc. 

Tables  containing  logarithms  of  more  places  can  be  pro- 
cured, but  this  table  will  serve  for  many  practical  uses,  and 
will  enable  the  student  to  use  tables  of  five-place,  seven- 
place,  and  ten-place  logarithms,  in  work  that  requires 
^eater  accuracy. 

In  working  with  a  four-place  table,  the  numbers  corre- 
sponding to  the  logarithms,  that  is,  the  antilogarithms,  as 
they  are  called,  may  be  carried  to  four  significant  digits. 

^       395.   To  Find  the  Logarithm  of  a  Number  in  this  Table. 

(1)  Suppose  it  is  required  to  find  the  logarithm  of  65.7. 
In  the  column  headed  "  N  "  look  for  the  first  two  significant 
figures,  and  at  the  top  of  the  table  for  the  third  significant 
figure.  In  the  line  with  65,  and  in  the  column  headed  7, 
is  seen  8176.  To  this  number  prefix  the  characteristic  and 
insert  the  decimal  point.     Thus, 

log  65.7  =  1.8176. 

(2)  Suppose  it  is  required  to  find  the  logarithm  of  20347. 
In  the  line  with  20,  and  in  the  column  headed  3,  is  seen 
3075 ;  also  in  the  line  with  20,  and  in  the  4  column,  is  seen 
3096,  and  the  difference  between  these  two  is  21.  The  dif- 
ference between  20300  and  20400  is  100,  and  the  difference 
between  20300  and  20347  is  47.  Hence,  ^^  of  21  =  10, 
nearly,  must  be  added  to  3075 ;  that  is, 

log  20347  =  4.3085. 

(3)  Suppose  it  is  required  to  find  the  logarithm  of 
0.0005076.  In  the  line  with  50,  and  in  the  7  column,  is 
seen  7050;  in  the  8  column,  7059:  the  difference  is  9.    The 


350  SCHOOL    ALGEBRA. 

difference  between  5070  and  5080  is  10,  and  the  difference 
between  5070  and  5076  is  6.  Hence,  y%  of  9  =  5  must  be 
added  to  7050 ;  that  is, 

log  0.0005076  =  6.7055  -  10. 

396.   To  Pind  a  Number  when  its  Logarithm  is  Given. 

(1)  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  1.9736. 

Look  for  9736  in  the  table.  In  the  column  headed  "  N," 
and  in  the  line  with  9736,  is  seen  94,  and  at  the  head  of 
the  column  in  which  9736  stands  is  seen  1.  Therefore, 
write  941,  and  insert  the  decimal  point  as  the  characteristic 
directs;  that  is,  the  number  required  is  94.1. 

(2)  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  3.7936. 

Look  for  7936  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
7931  and  7938 ;  their  difference  is  7,  and  the  difference  be- 
tween 7931  and  7936  is  5.  Therefore,  f  of  the  difference 
between  the  numbers  corresponding  to  the  mantissas,  7931 
and  7938,  must  be  added  to  the  number  corresponding  to 
the  mantissa  7931. 

The  number  corresponding  to  the  mantissa  7938  is  6220. 

The  number  corresponding  to  the  mantissa  7931  is  6210. 

The  difference  between  these  numbers  is  10, 
and  6210  +  ^  of  10  =  6217. 

Therefore,  the  number  required  is  6217. 

(3)  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  7.3882-10. 

Look  for  3882  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
3874  and  3892 ;  the  difference  between  the  two  mantissas 
is  18,  and  the  difference  between  3874  and  the  given  man- 
tissa 3882  is  8. 


LOGARITHMS.  351 

The  number  corresponding  to  the  mantissa  3892  is  2450. 
The  number  corresponding  to  the  mantissa  3874  is  2440. 
The  difference  between  these  numbers  is  10, 
and  2440  +  y\  of  10  =  2444. 

Therefore,  the  number  required  is  0.002444. 

Exercise  116. 
Find,  from  the  table,  the  common  logarithms  of  : 

1.  50.  4.    7803.  7.    7063.  10.   0.5234. 

2.  201.  5.    4325.  8.    1202.  11.   0.01423. 

3.  888.  6.    8109.  9.    0.00789.        12.    0.1987. 

Find  antilogarithms  to  the  following  common  logarithms: 

13.  4.1432.  15.    2.3177.  17.    9.0380-10. 

14.  3.5317.  16.    1.3709.  18.    9.9204-10. 

397.   Examples. 

(1)  Find  the  product  of  908.4  X  0.05392  X  2.117. 

log        908.4  =  2.9583 

log    0.05392  =  8.7318-10 

log       2.117  =  0.3257 

2.0158  =  log  103.7.  Ans. 

"When  any  of  the  factors  are  negative,  find  their  logarithms  with- 
out regard  to  the  signs  ;  write  —  after  the  logarithm  that  corresponds 
to  a  negative  number.  If  the  number  of  logarithms  so  marked  is 
odd,  the  product  is  negative ;  if  even,  the  product  is  positive. 

rn\    J?-   A  ,\.  r     ^.    f -8.3709x834.637 

(2)  l^ind  the  quotient  of ^^^^  ^.^ 

^  ^  ^  7308.946 

log      8.3709  =  0.9227 
log    834.637  =  2.9215  + 

colog  7308.946  =  6.1362-10  + 

9.9804  -  10  =  log  -  0.9558.  Am. 


352  SCHOOL    ALGEBRA. 

(3)  Find  the  cube  of  0.0497. 

log  0.0497  -  8.6964  -  10 
Multiply  by  3,  3 

6.0892  -  10  =  log  0.0001228.    Ans. 

(4)  Find  tbe  fourth  root  of  0.00862. 

log    0.00862=    7.9355-10 
Add  30 -30,  30.  -30 

Divide  by  4,  4)37.9355-40 

9.4839  -  10  =  log  0.3047.     Ans. 


(5)    Find  the  value  of  Ai^MlixiIIL21x2^18M.    ' 
^30.103^x0.4343^x69.897* 

log  3.1416=  0.4971  =0.4971 

log  4771.21=  3.6786  =3.6786 

^log  2.7183=  0.4343-2      =0.2172 

4  colog  30.103  =  4(8.5214  -  10)  =  4.0856  -  10 

^colog  0.4343=  0.3622-2      =0.1811 

4  colog  69.897  =  4(8.1555  -  10)  =  2.6220  -  10 

11.2816-20 

30  -30 

5)41.2816^ 

8.2563  -  10 

=  log  0.01804.  Ans, 

398.  An  exponential  equation,  that  is,  an  equation  in  which 
the  exponent  involves  the  unknown  number,  is  easily  solved 
by  Logarithms. 

Ex.    Find  the  value  of  x  in  8P  =  10. 
81*  =  10. 
.-.  log  (81*)  =  log  10, 
■   X  log  81  =  log  10, 

^  ^lo^^LOOOO^  0.524.  ^n,. 
log  81      1.9085 


LOGARITHMS. 


353 


Exercise  117. 
Find  by  logarithms  the  following  products 


1. 

948.7  X  0.04387. 

5. 

7564  X  (-  0.003764). 

2. 

3.409  X  0.008763. 

6. 

3.764  x(- 0.08349). 

3. 

830.7  X  0.0003769. 

7. 

-  5.845  x(- 0.00178). 

4. 

8.439  X  0.9827. 

8. 

-  8945.7  X  73.84. 

r 

'ind  by  logarithms : 

9 

7065 
5401 

11. 

0.07654 

83.94  X  0.8395 

10. 

7.652 
-  0.06875 

12. 

212  X  (-6.12)  X  (-2008) 
365  x(- 531)  X  2.576 

13. 

0.17681             17. 

my- 

21.    2.563TT 

14. 

1.2ir°.               18. 

906.8i 

22.    (8|)^-». 

15. 

Ill                    19. 

(mr- 

23.    (5||)°•"^ 

16. 

(Mr.                 20. 

(7Ar 

24.    (9H)l 

25.      »/0.0075^X 

78.34  X 

172.4i  X  0.00052 

4285i  X  54.27*  X  0.001  x  86.79^ 


26. 


27, 


f- 


0327P  X  3.429  X  0.7752=* 


32.79  X  0.00371* 


4 


7.126  xV0.1327x  0.05738 


VO.4346  X  17.38  X  VO.006372 


Find  X  from  the  equations : 

28.  5^=10.  30.    7*  =  40.  32.    (0.4)"*  =-3. 

29.  4^  =  20.  31.    (1.3)' =  4.2.      33.    (0.9)- =  2. 


354 


SCHOOL    ALGEBRA. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 
13 
14 

0000 
0414 
0792 
1139 
1461 

0043 
0453 
0828 
1173 
1492 

0086 
0492 
0864 
1206 
1523 

0128 
0531 
0899 
1239 
1553 

0170 
0569 
0934 
1271 
1584 

0212 
0607 
0969 
1303 
1614 

0253 
0645 
1004 
1335 
1644 

0294 
0682 
1038 
1367 
1673 

0334 
0719 
1072 
1399 
1703 

0374 
0755 
1106 
1430 
1732 

15 

16 
17 
18 
19 

1761 
2041 
2304 
2553 

2788 

1790 
2068 
2330 
2577 
2810 

1818 
2095 
2355 
2601 
2833 

1847 
2122 
2380 
2625 
2856 

1875 
2148 
2405 
2643 
2878 

1903 
2175 
2430 
2672 
2900 

1931 

2201 
2455 
2695 
2923 

1959 

2227 
2480 
2718 
2945 

1987 
2253 
2504 
2742 
2967 

2014 
2279 
2529 
2765 
2989 

20 

21 
22 
23 
24 

3010 
3222 
3424 
3617 
3802 

3032 
3243 
3444 
3636 
3820 

3054 
3263 
3464 
3655 
3838 

3075 
3284 
3483 
3674 
3856 

3096 
3304 
3502 
3692 
3874 

3118 
3324 
3522 
3711 

3892 

3139 
3345 
3541 
3729 
3909 

3160 
3365 
3560 
3747 
3927 

3181 
3385 
3579 
3766 
3945 

3201 
3404 
3598 
3784 
3962 

25 

26 
27 
28 
29 

30 

31 

32 
33 
34 

35 

36 
37 
38 
39 

3979 
4150 
4314 
4472 
4624 

3997 
4166 
4330 
4487 
4639 

4014 
4183 
4346 
4502 
4654 

4031 
4200 
4362 
4518 
4669 

4048 
4216 
4378 
4533 
4683 

4065 
4232 
4393 
4548 
4698 

4082 
4249 
4409 
4564 
4713 

4099 

4265 
4425 
4579 

4728 

4116 
4281 
4440 
4594 
4742 

4133 

4298 
4456 
4609 
4757 

4771 
4914 
5051 
5185 
5315 

4786 
4928 
5065 
5198 
5328 

4800 
4942 
5079 
5211 
5340 

4814 
4955 
5092 
5224 
5353 

4829 
4969 
5105 
5237 
5366 

4843 
4983 
5119 

5250 
5378 

4857 
4997 
5132 
5263 
5391 

4871 
5011 
5145 
5276 
5403 

4886 
5024 
5159 
5289 
5416 

4900 
5038 
5172 
5302 
5428< 

5441 
5563 
56S2 
5798 
5911 

5453 
5575 
5694 
5809 
5922 

5465 
5587 
5705 
5821 
5933 

6042 
6149 
6253 
6355 
6454 

5478 
5599 
5717 
5832 
5944 

6053 
6160 
6263 
6365 
6464 

5490 
5611 
5729 
5843 
5955 

6064 
6170 
6274 
6375 
6474 

5502 
5623 
5740 
5855 
5966 

5514 
5635 
5752 
5866 
5977 

5527 
5647 
5763 

5877 
5988 

5539 
5658 
5775 
5888 
5999 

5551 
5670 
5786 
5899 
6010 

40 

41 
42 
43 
44 

45 

46 
47 
48 
49 

6021 
6128 
6232 
6335 
6435 

6031 
6138 
6243 
6345 
6444 

6075 
6180 
6284 
6385 
6484 

6580 
6675 
6767 
6857 
6946 

6085 
6191 
6294 
6395 
6493 

6590 
6684 
6776 
6866 
6955 

6096 
6201 
6304 
6405 
6503 

6107 
6212 
6314 
6415 
6513 

6117 

6222 
6325 
6425 
6522 

6618 
6712 
6803 
6893 
6981 

6532 
6628 
6721 
6812 
6902 

6542 
6637 
6730 
6821 
6911 

6551 
6646 
6739 
6830 
6920 

6561 
6656 
6749 
6839 
6928 

6571 
6665 
6758 
6848 
6937 

6599 
6693- 
6785 
6875 
6964 

6609 
6702 
6794 
6884 
6972 

50 

51 
52 
53 
64 

6990 
7076 
7160 
7243 
7324 

6998 
7084 
7168 
7251 
7332 

7007 
7093 
7177 
7259 
7340 

7016 
7101 

7185 
7267 
7348 

7024 
7110 
7193 
7275 
7356 

7033 
7118 

7202 
7284 
7364 

7042 
7126 
7210 
7292 
7372 

7050 
7135 

7218 
7300 
7380 

7059 
7143 
7226 
7308 

7388 

7067 
7152 
7235 
7316 
7396 

LOGARITHMS. 

355 

N 

0 

1 

2 

3  1  4 

5 

6 

7 

8 

9 

55 

56 
57 
58 
59 

7404 
7482 
7559 
7634 
7709 

7412 
7490 
7566 
7642 
7716 

7419 
7497 
7574 
7649 
7723 

7427 
7505 
7582 
7657 
7731 

7435 
7513 
7589 
7664 

7738 

7443 
7520 
7597 
7672 
7745 

7451 
7528 
7604 
7679 
7752 

7459 
7536 
7612 
7686 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7551 
7627 
7701 
7774 

60 

61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 

7868 
7938 
8007 
8075 

7803 

7875 
7945 
8014 
8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 
8028 
8096 

7825 
7896 
7966 
8035 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8055 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 
8388 

8136 
8202 
8267 
8331 
8395 

8142 
8209 
8274 
8338 
8401 

8149 
8215 
8280 
8344 
8407 

8156 
8222 
8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 

8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8445 

70 

71 
72 
73 
74 

8451 
8513 
8573 
8633 
8692 

8457 
8519 
8579 
8639 
8698 

8463 
8525 
8585 
8645 
8704 

8470 
8531 
8591 
8651 
8710 

8476 
8537 
8597 
8657 
8716 

8482 
8543 
8603 
8663 

8722 

8488 
8549 
8609 
8669 

8727 

8494 
8555 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8745 

75 

76 
77 
78 
79 

8751 
8808 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8825 
8882 
8938 
8993 

8774 
8831 
8887 
8943 
8998 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8954 
9009 

8791 
8848 
8904 
8960 
9015 

8797 
8854 
8910 
8965 
9020 

8802 
8859 
8915 
8971 
9025 

80 

81 
82 
83 
84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 
9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9154 
9206 
9258 

9053 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9175 
9227 
9279 

9074 
9128 
9180 
9232 
9284 

9079 
9133 
9186 
9238 
9289 

85 

86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9400 
9450 
9499 

9301 
9355 
9405 
9455 
9504 

9309 
9360 
9410 
9460 
9509 

9315 
9365 
9415 
9465 
9513 

9320 
9370 
9420 
9469 
9518 

9325 
9375 
9425 
9474 
9523 

9330 
9380 
9430 
9479 

9528 

9335 
9385 
9435 
9484 
9533 

9340 
9390 
9440 
9489 
9538 

9586 
9633 
9680 
9727 
9773 

90 

91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 
9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 
9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9576 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

95 

96 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 
9974 

9800 
9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9854 
9899 
9943 
9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9952 
9996 

CHAPTER  XXVI. 
GENERAL    REVIEW^   EXERCISE. 
If  a  =  6,  5  =  5,  c  =  —  4,  d=  —  S,  find  the  value  of 
1.    Vb'  +  ac+-Vc'-2ac.         3.    'Slb^-j-ac+'Vc^ -2ac. 

Find  the  value  of 

^     X  .  X      1  ahc 

6.   -4-T>  when  ^ 


a      6  a-\-b 

6.  -(5  — :r)  +  -(c  — :r)  +  -(rf:— c),  when  a;= — — — — — 
aba  a 

„     X  .       X  1  d^ib  —  a) 

7.  --}-- ,  when  x  =  —^^ -^• 

a     b  —  a  D{o-\-a) 

8.  (a -\- x)  (b -\-  x)  —  a  (b  -\-  c)  -\-  x"^,  when  x  =  -—' 

0 

9.  — ^^ — '    /   ' — ,  when  x  =  —  a. 

a{l-]-b)  —  bx     a  —  2bx 

10.  Add{a-b)x'-{-(b-c)fi-(c-a)z\(b-c)x^+(c-a)^ 

+  (a-b)z\  and  (c  -  a)x'  +  (a-  b)y'  +  ib  -  c)z\. 

11.  Add  (a  +  5)a;  +  (5  +  c)y-(c+a)z,  (b-\-c)z-{-(c-{-a)x 

~-{a-\-b)y,  and  (a  + c)y  +  (a  +  5)z— (5  +  c)ar. 

12.  Show  that  a;'  +  2/'  +  z'-3a;yz  =  0,  if  a;  +  y  +  z  =  0. 
J3.   Show  thata;'-8/-272'-18a;yz  =  0,  iix=2y-\-Zz. 


GENERAL    REVIEW   EXERCISE.  357 

Simplify  by  removing  parentheses  and  collecting  terms : 

4.  Ba^2(b-c)-[2(a-b)-^(c+a)]-[9c-4:(c-a)]. 

15.  7(2a  +  b)  -  \19b  -  [13((7  -  a)  +  12(6  -  c)]\. 

16.  a:-{4y  +  [3(z-a;)-(2:+2y)]-(2y  +  z-2a:)}. 

17.  l  +  2{a;  +  4-3[a;+5-4(a;+l)]|. 

18.  10a7-^;4[5a:-3(a:-l)]-3[4^-3(a:  +  l)]5. 

19.  ?>x'-\2x'-{?>x-1)-[2x'-{Bx-x'y\-[b-{2x'-^x)]\. 

20.  (a;  -  2)(a;  -  3)  -  (a; -  7)(a:-  1)  +  (:r  -  l)(a:  -  2). 

21.  (;r  +  2/)^-2a:(3a;  +  2y)-(y-^)(-a:  +  y). 

22.  3a-[2a-(3a-^»7]  +  3a/'2^»-3a-— \ 

Resolve  into  lowest  factors  : 

23.  (x  +  yy  —  ^z\  27.    9x^ -^xY -\-^y\ 

24.  {x^  +  yj-4.xy.  28.    ("^ J%  ^IJ"*- 2. 

25.  a^-b^-c''-\-2bc.  29.    81a*— 1. 

26.  (rr'-y'-227-4yV.  30.    a^^-5". 

31.    (a2-52^c='-^7-(2a^-2^»^)'. 

32.  :i;'— 19a:-f  84.  40.  a:'-y«. 

33.  Ja;2+ 2J:r  — 36.  41.  8  +  aV. 

34.  a:'  — 8a; +15.  42.  3^  -  a}\ 

35.  9a;'  — 150 a: +  600.  43.  27a' -64. 

36.  56a;^  +  3a;y-20y^  44.  a,-^»-32y*. 

37.  12a;^  +  37a;  +  21.  45.  a}^-b\ 

38.  33 -14a; -40a:'.  46.  a;'*  +  1024y»<>. 

39.  6a:' +  5a; -4.  47.  a^-{b-\-cf. 

48.    8ar*-6a;y(2a;+3y)  +  27y'. 


358  SCHOOL    ALGEBRA. 

Find  the  H.  C.  F.  of 

49.  6x*-2x'  +  9x'-\-9x-4:,  and  9a;*  +  80a;'  — 9. 

50.  3a;^-5a;'  +  2,  and  2a^-5x'  +  S. 

51.  a;' -93a; -308,  and  a;' -  21a;' +  131a;  — 231. 

52.  a;*  — 2a;'+4a;'-6a;+3,  and  a;*  — 2a;'  — 2a;'+ 6a;-  3. 

53.  a;^-4a:'-a;'  +  2a;  +  2,  and  a;'- a;' —2a;  +  2. 

54.  3a;' +  10a;' +  7a; -2,  and  3r' +  13.^'+ 17a;4- 6. 

55.  4a:*-9a;'  +  6a;-l,  and  6a;'-7a;'  +  l. 

56.  a;^+lla;-12,  and  x' -}- 11  x^ -}- 64:. 

Find  the  L. CM.  of 

57.  4a;' +  4a;  -  3,  and  4a;' +  2a; -6. 

58.  a;'  — 4?/',  and  or'  +  ^y  — 6y^. 

59.  7a'a;(a  — a;),  21  aa;  (a' —  a:'),  and  12ax^(a-]-  x). 

60.  9a;' -a; -2,  and  3a;' -  10a;' -  7  a;  — 4. 

61.  a;'-5a;  +  6,  a;'-4a;+3,  and  a:'  — 3a;  +  2. 

62.  2a;'  +  bx^y  —  5xy^  +  y\  and  2a;'  —  7a;'y  +  5xy^  —  y'. 

Simplify : 

63.  ^4-1  3a;  +  2    ^      2a;-l 


a;(a;-2)      a;(a;+l)      a;'-a;-2 

64     1  +  ^  ,  1-^      1  +  ^'      1-^ 
1-a;      1  +  a;      1  -  a;'      1  +  a;'* 

gg  a;-l 2(a;  +  2)        .  a;  +  5 

•    (^  +  2)(a;  +  5)      (a;  +  5)  (a;  -  1)  "^  (a;  -  1) (a;  +  2)* 

66.    -i_,-f  1 


ax  —  d^      aa;4-2a'      a;'  +  aa:— 2a' 
g^  a; L  a;  — 1  a;  — 3 


(^+3)(a;-l)      (a;  4- 3)  (2 -a;)      (2-a;)(l-a;) 


GENERAL    REVIEW   EXERCISE. 


359 


"■  ('+.4T+^)(>+.-i,-:?-3) 

69.    f^  +  ^y  I  A  :   f    ^  ^  +  2?A 

a^-5^ «-^  l/a  +  5 1_\ 


71. 


72. 


a      0      c 


1  —  a      1  +  a 

a       ,       1 


1 —a      1+a 


74. 


75. 


76. 


5-1  + 


a;-l  + 


1  + 


4:  —  X 


1  + 


l  +  «  + 


2a^ 


1+a 


Solve : 


6a;+13      9:c+15  .  ^_2x-\-\b 
15  5a;-25"^  5 


2a:  +  a     ■     ^x-a  _oi 
•    3(a;-a)'^2(a;  +  a)      ^' 


79. 


a:4-l      ^—8      a:-9 


—  2      a;— 1      07—6      a:  — 7 


80.    |+^=4i 
£+12  =  21 


81. 


0.7 

=  6 

10:r 
7 

=  31 

360 


SCHOOL   ALGEBRA. 


82.    2£^3^_4._^ 
a        0        c 

a        h        c 

a        h        c 


83. 


X     y      z 


a      h 
X     y 


2a      b     c 


=  3 

=  1 
=  0 


Find  the  arithmetical  value  of 

84.  36^;  27^;  16^;  32^;  4^  8^ ;  27^;  64^ 

85.  32';  64^  81^  (3f )^ ;  (S^V)^ ;  {l^\)\ 

86.  (0.25)^  (0.027)^  49»-^;  32"-^;  81° ^ 

87.  36"^  27-^(yy~^  (0.16)-^  (0.0016)"l 

88.  Interpret  a"';  a°;  a^ ;  a~^;  (a"^)~i 
Simplify : 

89.  a^Xa^;  c^  X  c^  ;  m^  X  m~^  ;  n^  X  n'^^. 

90.  a^^y  X  a~^6''3c""^  ;  ahh~^  X  a^5~yc?. 

91.  (2ab  +  2bc-\-2ac-a'-b'-  c')  -^  (J  +  b^  +  c^). 

92.  Vl2;  V8;  VSO ;   ^/U]  4^250;   Vl;   -v/i;   V^. 

93.  b^/^^MO;  -V^P]  -^/'a']  Vd'x  +  a' ;  3J■^54i■^ 

94.  2V18-3V8  +  2V50;  "v/Sl  + -^24  -  \/l92. 

95.  fVf+ V80-iV20;  8VS+10V||-2V|. 


96. 


97. 


Kationalize  the  divisor,  and  find  the  value  of 
1  ..  5 


2-V3 

3 

3  +  V6* 


98. 

V2  +  V7 

99.    V3  +  V2 
V3-V2' 


100. 


101. 


7V5 


V5  + V7 
2V6 

V3  -  Vs' 


O^KERAL   review   EXERCiSfi.  861 

Solve : 

109.  _i_  +  _X_  =  ^. 

110.  ax"" ^-^  =  cx-hx'. 


102. 

5         4         3 

x-2     X     x+Q 

103. 

a:  +  3       2x-l_^ 
2x-l       x-?> 

104. 

:-^t+si=«.. 

106. 

x  —  a  .  x—h_ c^-\- V 
x  —  b      x  —  a        ab 

106. 

ax -\- b  _  ex -\-  d 
a-\-bx      c  -\-  dx 

in7 

c             d         c  —  d 

X  —  d     x-\-  c         X 

108. 

ax     .      bx             ,7 

a-^b 

111.  £±«^^±1_2. 
x-\-b     x-\-c 

112.  3^2/-5y^=   n 
bxy-{-^x'  =  22) 

bxy-2,y'=    2) 


114  Va;  +  y=:Vy  +  2| 
:r-y  =  7  j 

115.  ^^  +  ^y  +  y'  =  52| 

X  —  ox  —  a  xy  —  x^  =  ^  J 

Form  the  equations   of  whicli  the  roots  are : 

116.  a  —  b,a  +  b.  119.  1  +  V3,  1  —  VS. 

117.  a  — 25,  a  +  35.  120.  -1  +  V3,  -1-V3. 

118.  a  +  25,  2a  +  i.  121.  1  +  V^^,  1-V^. 

Solve : 

122.  :r*-5ar'  +  4  =  0.  127.  2ar«- IQrr' +  24  =  0. 

123.  :r«-9a;'  +  8  =  0.  128.  or*- 1  =  0. 

124.  9:r*-13a:'  +  4  =  0.  129.  a:«-l  =  0. 

125.  4rc*-17a:*  +  4  =  0.  130.  rc^  +  8a;^  -  9  =  0. 

126.  2a;*-5a^  +  2  =  0.  131.  16x^  -  17a:*+ 1  =0. 


362 


SCHOOL   ALGEBRA.. 


132.  2^;^  — 3a;^+l  =  0. 

133.  S</x-b</x  +  2  =  0. 

134.  6V^-3\/^-45=0. 

135.  21\/?-5v'i-74  =  0. 

136.  3V?4-4\/^-20  =  0. 

142.  3\/9^+4-{/3^-39  =  0. 

143.  V3a^  +  a\/3a^  — 2a'  =  0. 

144.  ^■\/2b^-5b</2b^-2b'  =  0. 


137.    x'^  +  bx'^-U^O. 


138.    4a:~*-3a;"^-27  =  0. 


139.  a;''*  +  3a;"  — 4  =  0. 

140.  Sx^-2ax^  —  a''=0. 

141.  V2a;--\/2a:-2  =  0. 


145.    Vx-\-4:  +  -VSx+l  =  -\/9x  +  4: 
146. 


5a:  +  1  +  2V4:r  -  3  =10  V^  -  2. 
147.    2Vx  +  2  --  3  V3a;  -  5  +  VSa;  +  1  =  0. 


148.    Vll  -  a;  +  V8  -  2a;  -  V21  +  2a;  =  0. 
Expand : 

149.  (J-x^y-,  (Sa^-2bJ;  (J-ih^Y;  (2  -fj;  f^-fj' 

150.  ra;'_^Y;  (2a;'+V3^)*;  (V^l-\-aJ;{l+2x-x'-3^y. 

151.  Expand  to  four  terms 

(l-3a:)-^;  (l-4a;')-^  (1-fa:^)^;  (a-2x-^)-\ 

152.  Find  the  eighty-seventh  term  of  (2  a;  —  y)^. 

153.  Resolve  into  partial  fractions 

3 -2a:  3 -2a;  1 

l-3a;  +  2a;''    (1  -  a;)(l  -  3a;)  '    l-a^' 

154.  Expand  to  five  terms --r- 

.  ^  l-3a;  +  2a:* 


■      or  THf     *^  \ 

UNIVERSITY 


/Uv^^ 


7  ^+-'^-^^ 


^ 


^ 


4^^ 


^ 


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